We are Keeping It CLEAN and we welcome you! You will find we are the Maid Service in Atlanta that plays for keeps. We started with a small crew and small jobs and today can proudly say we are able to tackle the biggest of cleaning needs!
You can call Keeping It Clean for any cleaning job, be it Commercial or residential, not a job to small. You can call Keeping It Clean for a quote at 404-996-1129 or simply visit our quote page. If for any reason we get there and something was overlooked we will not start the job until you have an exact figure. No surprises from us unless a great one! No hidden cost we are always upfront. we are Bonded, Licensed and Insured and all employees must pass a drug screen and a nationwide criminal background check to be employed by Keeping It Clean.
count on. We realize that this is earned and will never take that for granted!
If you don't see listed what you are looking for, don't hesitate to visit our Quote Page to contact us with your request!


text:
50/50 Not Always Fair under Child Support Standards Act
By Friedman & Friedman, Attorneys at Law
When people go through a divorce, the law basically says whomever the child primarily lives with is the recipient of child support — and the amount the other person pays is based upon the Child Support Standards Act. Today, the courts need to realize that society has changed since the CSSA was enacted in 1989.
In today's society, more and more people are meeting at work and spouses are doing the same jobs and having equivalent income. Very often, when these couples file for divorce they seek 50/50 sharing of time with their children. I am seeing this come through my office more and more. I am finding people who are talking about joint legal custody who have the same jobs — including teachers, nurses, lawyers — and they have the same income.
What do you do about child support when they have the same job/income?
Although the CSSA has been periodically challenged — The New York Court of Appeals set the tone, in 1998 in Bast v. Rosoff, ruling that the Child Support Standards Act must be followed.
With continued change in the family dynamic, so far, the higher courts have not addressed a shared 50/50 formula. However, I have seen a number of trial courts deviate from CSSA when there is an equal sharing of parental time and very similar incomes. These trial courts are sending out a message that basically says, "Look, The Child Support Standards Act says that we don't have to apply it if we go through the mathematical calculation and find the act to be unjust or improper — and we can deviate."
It is time that the higher courts address this subject matter and establish that when they find the CSSA to be unjust or improper, adjustments need to be made.
Categories: February 16, 2015

redpajama_set_name: RedPajamaCommonCrawl


Caskinette Ford in Carthage, NY treats the needs of each individual customer with paramount concern. We know that you have high expectations, and as a car dealer we enjoy the challenge of meeting and exceeding those standards each and every time. Allow us to demonstrate our commitment to excellence!
Our experienced sales staff is eager to share its knowledge and enthusiasm with you. We encourage you to browse our online inventory, schedule a test drive and investigate financing options. You can also request more information about a vehicle using our online form or by calling (315) 493-1360.
Caskinette Ford is driven to ensure every customer with the highest levels of customer service, providing a car shopping experience like no other. For more than 40 years, Caskinette Lofink Ford, located in Carthage, NY, has been a intregral part of the Carthage NY,Fort Drum, Lowville, Watertown NY, and surrounding communities, offering the best inventory of new Ford F-150, Ford Escape, Ford Explorer, Ford Fusion, Ford Focus, Ford Fiesta and the entire Ford line of cars and trucks. We also continue to service those vehicles and are proud to maintain our customers' rides for the duration of ownership. And when the time comes to buy or lease, our automotive financing finance can help. Loyal customers continue to trade with us, giving us the best inventory of pre-owned used vehicles in Northern New York. We invite everyone to experience the Caskinette's difference. Call us today at 866-417-8973 and speak to someone on our team. Make Caskinette's your car dealership for years to come.
Your satisfaction behind the wheel is incredibly important to us here at Caskinette Ford, and our wide selection of new Ford models and used vehicles is sure to impress any and all drivers in the Carthage, NY area! Whether you've got a big family that's always on the go or you tend to drive solo, you need a large and in-charge powerhouse of a truck or you need something small and efficient, whether you're looking to drive through urban streets or you'd like to travel the forested trails, we'll have something to perfectly fit your driving desires. We welcome one and all to take a look through our lineups and find something that's as sleek and safe as it is fun and affordable.
One look through our selection of new Ford models and you very well may end up head-over-heels in love at every turn! We have a huge selection of new Ford vehicles of all types, sure to satisfy your specific wants and needs in the driver's seat; with popular models like the Ford C-MAX, EcoSport, Escape, Explorer, F-150, Focus, Fusion, Mustang, Super Duty, and many more, your preferred on-road companion could be just a click away. And with a list of new Ford model specials to explore, you could stretch your budget even further without breaking a sweat.
We know that not all Fort Drum drivers are looking for something brand-new and, if this describes your own interests, our selection of used vehicles and certified pre-owned Ford models are bound to make a strong impression. You'll find plenty of pre-owned Ford models in this selection as well as a multitude of other popular models from automakers like BMW, Chrysler, Dodge, GMC, Hyundai, Jeep, Kia, Nissan, Ram, Toyota, and more; whether it's a car, crossover, SUV, truck, or minivan that's on your mind, we'll have something here to pique your fascination. We even have a list of used vehicle specials as well as a whole selection of bargain pre-owned vehicles under $15,000 so you can save even more on your hunt for the right ride.
Finding a great vehicle is just the first step in the process of getting it home; if you're in need of a little financial assistance, the experts at our finance center are ready to offer their help. Whether you're looking to secure a loan or a Ford lease, our finance team will discuss various options and payment plans until you find the one that best fits your budget. If you'd like to get financial assistance from the comfort of your Lowville home, you can even apply for credit right on our website! Your satisfaction is our top priority.
After you've been driving around in your new vehicle for a while and you find it's time for some routine maintenance, or you've been in an accident and need repairs, you can easily contact our service center as well. After simply scheduling an appointment on our website, our technicians will get back to you with a time that best fits your days. Then, when you arrive, you can leave all the hard work in their professionally-trained hands. To save you even more money, don't forget to explore our current lists of service coupons and service specials, sure to keep your wallet fat and happy while you get the repairs and maintenance you demand.
There's a lot to discover at Caskinette Ford, so drop in at your earliest convenience and take advantage of everything we have to offer! If you have any questions for us, call us at 877-358-3914; for everything else, we're located conveniently in Carthage, NY and can't wait for your arrival. We look forward to serving you in the near future.
driving instructions to our site, or give us a call. We look forward to serving you!


text:
The women's 100 metres hurdles event at the 2020 Summer Olympics took place between 31 July and 2 August 2021 at the Japan National Stadium. 40 athletes from 28 nations competed. In the semifinals, Jasmine Camacho-Quinn of Puerto Rico broke the Olympic record, running 12.26 secs, to go equal fourth on the world all-time list. The following day in the final, she won the gold medal with a time of 12.37 secs. American world record holder Keni Harrison finished second to clinch silver and the bronze to Jamaica's Megan Tapper.

Summary
Jasmine Camacho-Quinn came into the Olympics with the top 3 times of the year, making her the one to beat. After failing to make it to the 2016 Olympics, Kendra Harrison set the world record as a consolation. Five years later, this was her first Olympics. In the semi final round, Camacho-Quinn confirmed her favorite status by setting the Olympic Record at 12.26, which also equalled the #4 time in history and set her national record. It took 12.67 just to get into the final.

From the gun, the final was a two-person race, Henderson out fast but Camacho-Quinn almost matching her for the first couple of hurdles. From there, Camacho-Quinn just was moving faster, passing over the fourth hurdle and pulling away on every hurdle after that. Out in lane 8, Megan Tapper was the next closest, slowly gaining on Harrison. Camacho-Quinn was the clear winner. Tapper couldn't catch Harrison for silver, but the judges spent several minutes confirming the photo finish before posting the next two places. The winning time was 12.37 as every athlete save Tapper and 4th place Tobi Amusan had run faster in the semifinals the previous evening.

Background
This was the 13th appearance of the event, having appeared at every Olympics since 1972.

Qualification

A National Olympic Committee (NOC) could enter up to 3 qualified athletes in the women's 100 metres hurdles event if all athletes meet the entry standard or qualify by ranking during the qualifying period. (The limit of 3 has been in place since the 1930 Olympic Congress.) The qualifying standard is 12.84 seconds. This standard was "set for the sole purpose of qualifying athletes with exceptional performances unable to qualify through the IAAF World Rankings pathway." The world rankings, based on the average of the best five results for the athlete over the qualifying period and weighted by the importance of the meet, will then be used to qualify athletes until the cap of 40 is reached.

The qualifying period was originally from 1 May 2019 to 29 June 2020. Due to the COVID-19 pandemic, the period was suspended from 6 April 2020 to 30 November 2020, with the end date extended to 29 June 2021. The world rankings period start date was also changed from 1 May 2019 to 30 June 2020; athletes who had met the qualifying standard during that time were still qualified, but those using world rankings would not be able to count performances during that time. The qualifying time standards could be obtained in various meets during the given period that have the approval of the IAAF. Only outdoor meets are eligible. The most recent Area Championships may be counted in the ranking, even if not during the qualifying period.

NOCs can also use their universality place—each NOC can enter one female athlete regardless of time if they had no female athletes meeting the entry standard for an athletics event—in the 100 metres hurdles.

Records
Prior to this competition, the existing world, Olympic, and area records were as follows.

The following record was established during the competition:

The following national records were established during the competition:

Schedule
All times are Japan Standard Time (UTC+9)

The women's 100 metres hurdles took place over three consecutive days.

Results

Round 1
Qualification Rules: First 4 in each heat (Q) and the next 4 fastest (q) advance to the semifinals.

Wind readings- Heat 1: +1.0 m/s; Heat 2: +0.4 m/s; Heat 3: +0.4 m/s; Heat 4: -1.1 m/s; Heat 5: +0.3 m/s

Heat 1

Heat 2

Heat 3

Heat 4

Heat 5

Semifinals
Qualification rules: First 2 in each heat (Q) and the next 2 fastest (q) advance to the finals.

Wind readings- Heat 1: -0.8 m/s; Heat 2: +0.0 m/s; Heat 3: -0.2 m/s

Semifinal 1

Semifinal 2

Semifinal 3

Final

Wind reading: -0.3 m/s

References

Women's 100 metres hurdles
2020
Women's events at the 2020 Summer Olympics
Olympics

redpajama_set_name: RedPajamaWikipedia


We've heard from many people that finding a good financial adviser or financial planner can be a daunting task. There are many pitfalls to watch out for and a myriad of options so how do you know who is right for you? We've tried to summarise the main things to think about and some useful tips for making the right choice.
What is a financial adviser / financial planner?
A financial adviser is someone who can help advise you on your money. This will usually cover the recommendation of financial products like investments, pensions, insurance and so on. A financial adviser has a professional responsibility to provide you with a financial solution to meet your needs.
A financial planner is similar to a financial adviser but will usually focus more on you and your goals. They will then recommend financial products where a need arises to help you meet your objectives.
While this distinction may seem small, in reality there is often a significant different between advisers and planners. Usually financial planners do provide full financial advice but usually focus more on you as a person, your goals and objectives before getting into a discussion about financial products.
When thinking about finding a financial adviser or financial planner, you should think about what you need from them and how they might be able to help. Are you looking for something simple like a review of your investments or do you want a more in-depth discussion about you and your money. If it's the latter, try to find a good financial planner. If it's the former, either a financial adviser or financial planner should be able to help.
From this point on, we'll simply refer to 'financial advisers' but everything in the rest of this article refers to both advisers and planners equally.
How do I know the adviser is legitimate?
Without doubt, the most important thing to check with any financial adviser is that they are registered with the Financial Conduct Authority (FCA). This is the UK financial regulator and all advisers should be registered. You can look at the FCA register to check that the firm you are looking at is registered, meaning they are authorised to give advice and have passed a number of background checks. This includes having appropriate qualifications, professional standing and capital reserves. Most advisers will have a statement such as 'Authorised by the Financial Conduct Authority' on their website and stationery and some may quote their authorisation number. It is prudent to check this on the FCA register before engaging their services.
All advisers also need to have an annual 'Statement of Professional Standing' (SPS) certificate, issued by an FCA-accredited body. This will attest to their qualifications and continuing professional development and their inclusion on the FCA's register. If you're in any doubt, request to see the SPS for the adviser you are considering working with.
Should I opt for an independent or restricted adviser?
There are two types of financial adviser. 'Independent' advisers are obliged to consider all products and services and cover the full range of financial products. 'Restricted' advisers may only recommend the products of a particular company or have a limited number of companies they work with. Another reason some advisers may be 'restricted' is that they only offer advice on particular areas such as pensions or investments and don't cover the full range of financial products. Be careful because restricted advisers don't have to advertise that they are 'restricted' upfront, but they do have an obligation to explain their restriction before you sign-up to become a client. Only independent advisers can use the term independent, but again, care is needed as words like 'impartial' or 'unbiased' are often used in the promotional materials of both independent and restricted advisers.
An independent financial adviser is generally seen as the best option here, but the additional work involved in being independent could mean that charges may be higher for this type of service.
All regulated financial advisers must have a basic level of qualification. This is known as 'Level-4'. Some advisers have now achieved 'Level-6' and are known as either a 'Chartered Financial Planner' or 'Certified Financial Planner'. There are a myriad of different qualifications concerning specialist areas of advice like mortgages, equity release and so on, but it is this core level of qualification that you should look for. Ask if the financial planner you are considering is 'Chartered' or 'Certified' and if not, Google their qualification and check them out.
For a detailed list of financial adviser qualifications, check out Unbiased qualifications.
You should think about the charging method you are most comfortable with before selecting a suitable financial adviser. No one method is necessarily superior to any other, but each has pros and cons.
Am I free to start and stop charging as I like or am I tied into a contract?
Are the charges 'all in' meaning I might be paying for services I don't actually use?
Are the charges clear and transparent or are there likely to be hidden costs along the way?
Are the charges shown upfront on the adviser's literature and website – and if not, why not?
What extra services will the adviser provide me?
Do I like the look and tone of the firm and/or adviser?
Just like with any company you choose to use, you need to make sure the financial adviser you work with looks and acts the part and they fit with what you are looking for. You might look at the quality of their website, their business card, social media or their personal appearance. All these things together will help you form an image of the company and adviser you are considering. Are they professional, stuffy, modern, interesting, staid, etc. Again, there is no right and wrong, but you need to find an adviser that's a good fit for you and one that you are impressed with when you look at their profile or meet them in person.
What do other people say about this adviser?
Nowadays, most people check out the reviews before they buy pretty much anything and the same is true when selecting a financial adviser. Websites like VouchedFor and Google can be a great place to start. Increasingly, more advisory firms are conducting customer satisfaction surveys and publishing their results as we do. Most adviser website will also feature testimonials from happy clients and some firms may let you speak to these clients to validate the quality of their services.
However, probably the best source of reviews is people you know. If someone you respect rates a particular adviser, this often goes a long way to ensuring they will offer you a great service. This is where schemes like our own 'Recommend A Friend' scheme are useful for existing clients to refer new clients who could benefit from the services of a particular financial adviser.
Can I speak to the adviser before signing up for their services?
Most advisory firms now offer a free introductory meeting either face-to-face, via the phone or online. However this is done, it's a great opportunity to question the adviser and hear how they think they can help you. It's important to listen to the tone and way things are said as well as to the content. You need to like what you hear and imagine yourself working with the adviser for years to come. If you can't, then they probably aren't right for you.
Where should I look for a great financial adviser?
We hope that you have found this guide to finding a financial adviser useful and that you feel empowered to find one to perfectly suits your needs. At NorthStar, we believe that more people should be able to access great quality financial advice, financial planning and wealth management and hope this guide can help in a small way to further this cause.


Our Dream Center on North Church street is a hub of activity for our community. We partner with community groups and provide services for the whole family and all ages of children and students. We have a large meeting space, classrooms, a dance studio and full kitchen along with a running track, computer lab, and giant warehouse that is being converted to an indoor soccer field, sport court and coffee bar. We strive to provide a fun, creative, and safe environment to encourage dreams and see destinies transformed.
Spin-Weekly middle school age ministry 6-8pm Wednesday nights for students in the Broadview area. Dinner together, fun, small groups led by adult and high school student leaders, encouraging message and more. We provide transportation for about 30 of our students each week. Our weekly attendance is 45-70 students.
El Centro-Our bilingual family celebration. A fun casual evening for the whole family on the 3rd Saturday of the month 5-7pm. We have fun activities for all ages and share a meal together.
Un Grupo Comunitario-La Vida junta tiene un color especial.
A community group– Life together has a special color.
1st Sunday of the month 5-7pm Worship and classes for adults, students, and children.
reunión del segundo y cuarto sábado 5-6: 30. adoración, mensaje y compañerismo en español.
Our outreach ministry is a growing expression of the vision of CityGate Church. Our purpose is to Connect people with God. We believe the Lord's heart beats to reach the lost and hurting and we are endeavoring to reach out across our community to serve in many practical and tangible ways. If you have a heart to serve and would like to join us take a look at our current outreach opportunities and get as involved as you would like to be. It could be as simple as donating to our snack pack collections or helping us lead one of our monthly areas of ministry. More doors are continuing to open, Join your dream to ours and lets see how the Lord leads us in this growing adventure.
For more information contact lisa@citygatechurch.com or call the church office 336-586-CITY (2489).
Shine Camp – Shine Camp is 5 fun filled days during the summer for the students of Andrews Elementary School. A combination of adults and middle and high school students commit to making this the best week of summer for all the children who attend. After our transportation and breakfast teams have done their work, we kick off a fast paced and exciting program including music, dancing, acting, games, small groups, and a great message. The kids then get to choose 2 break out sessions like crafts, dance, soccer, games and so much more. The commitment of our volunteer staff is to focus on each child and to go above and beyond in making each day packed with smiles and lots of laughter.
Saturday Shine – On the 4th Saturday of each month we take over the gym at Andrews Elementary school and fill the day with fun and special activities for the 1st-5th graders. We provide 2 meals,transportation, crafts, fun, activities, sports, and a lively celebration each month for 100+ children. If you would like to join our team meet us at The CityGate Dream Center at 9 am. We begin set up and transportation at 9 and run our program 10am-2pm.
Adopt-A-School – Andrews Elementary Snack Packs-Providing 250 monthly snack packs, bags of easy to prepare food students can take home with them for the weekend. We assemble the snack packs on the 1st and 3rd Wednesday in the foyer. Or if you would like to donate toward the needed items pick up a shopping list at the display in the auditorium. Quarterly Celebrations-Providing support, volunteers and funding for quarterly reward celebrations. We also help sponsor dances and fundraisers to help with the 5th grade trip.
Love My City Giveaway– Twice a year we provide a giant giveaway to the community. Clothing, shoes, housewares, toys, furniture, food and more. Donate your gently used items. We collect donations on the second Sunday of the month.
Thanksgiving and Christmas Projects – Thanksgiving Turkey Dinners-We provide complete ready to prepare meals to families needing assistance through the school and our community connections. Last year, through generous contributions.
Thanksgiving Meal-The Sunday before Thanksgiving we will be serving a Thanksgiving meal to our Dream Center community. You can help cook, bring a food donation or come and serve.
Beary Christmas- Door to Door delivery of new cuddly teddy bears to area children. We provide 250 per year. It is a great outreach for the whole family and will become a highlight to your Christmas season. Bring your new teddy bear to the church and we will tag it and prepare it for our delivery.
Angel Tree – Through our school connection our people were able to bless children with clothing and toys for Christmas. Each child receives 5 gifts. The parents have the joy of giving their children the gifts and seeing their eyes light up on Christmas morning. You can sponsor one of the gifts for a child or sponsor a child for all 5 gifts. We receive the names and sizes through our partnership with Andrews Elementary and the gifts are given to the parents through the school.
Senior Care – The first and fourth Thursdays we meet at the Alamance Healthcare Center and sing hymns and encourage the residents from 10-11am Join us and be a blessing to a community that is often overlooked.
Sports Camp-4 day sports camp run in 2 tracks Students 6-10th grade and elementary children 1st-5th grade. We currently offer gymnastics, soccer, basketball, volleyball, track. This camp is free for all participants and includes 2 meals, instruction and transportation for those within the Citygate dream center community. We meet at the dream center and then transport the participants to nearby parks and fields. This summer we had 150 kids and teenagers participate and due to Sponsorship from Soccer.com were able to give each participant soccer gear and footwear.
Performing Arts Camp-4 day performing arts camp run in 2 tracks Students 6-10th grade and elementary children 1st-5th grade. . We currently offer acting, choir, instruments, contemporary dance, hip hop dance, and art. This camp is free for all participants and includes 2 meals, instruction and transportation for those within the Citygate dream center community.


For anyone dealing with mental health disorders, the ability to manage medication is essential to wellness. Adults in poor physical health, and who consequently have multiple prescriptions, are more likely to struggle with taking their medications appropriately and effectively. Sometimes, medications prescribed for medical conditions can exacerbate or even cause behavioral symptoms.
To help you achieve optimal wellness, Haven Behavioral Hospital of Eastern Pennsylvania offers a medication management program designed to ensure you are getting the best possible outcomes from your treatment.
Our board-certified clinicians will work with you and your other doctors to coordinate proper dosages, eliminate redundancies and ensure there are no potentially negative drug combinations. Together, our expert staff will work to refine your medication regimen to provide the best solutions from both a psychiatric and medical perspective. Our goal is to keep medication simple — the fewer and more effective the medications you take, the safer and healthier you will be.


Industry reporter Mary Jo Foley confirmed on her Twitter that the Windows 10 Anniversary Update ISO will be available on August 2. She had a conversation with a Microsoft spokesperson who assured her that there wouldn't be any problems with the availability of the Anniversary Update ISO on the same day when the AU will be released.
Many users were skeptical that Microsoft would shift the release date at the last minute and won't keep its word about the release of the Anniversary Update on August 2. Today is the last day Windows 7, 8 and 8.1 users are allowed to upgrade to Windows 10 for free if they purchased a valid license for their current operating system. Those who took this step are now waiting to get their hands on the Windows 10 Anniversary Update, which will be released in just four days.
According to Microsoft, the AU will be released in stages and will eventually reach all 350 million devices in the near future. Still, there are users wondering if it will be possible to download the ISO of the major release on the same day. After talking to a Microsoft spokesperson, Mary Jo Foley posted on her Twitter account, saying "For those asking, MS spokesperson just told me Windows 10 Anniversary Update ISOs will be available from MS on Aug. 2, too".
The Windows 10 Anniversary Update ISO will be useful for users who wish to proceed with a new installation of the operating system. Also, by installing the AU via ISO, they will overcome the predictable congestion generated by the Windows Update service, so they will be able to download the new update on August 2.
Windows 10 Anniversary Update will include several new features like Windows Ink, which will allow Surface owners to interact better with the tablet's interface using the stylus, an upgraded and smarter Cortana, extension support for Microsoft Edge, and the transformation of Skype into a universal app.


Just finished up a wedding commission for a couple based in Sligo, Ireland capturing a beautiful fleeting moment and making it permanent. This was the perfect gift to remember their big day from.
How can we represent and give justice to the fruitful creative minds of Irish culture?
For me it starts with colour, shapes and texture. An amalgamation of these things with a touch of spontaneity came together to produce this Jackson Pollock-like painting of Seamus Heaney, an Irish poet and playwright. I wanted to capture him deep in introspection, philosophising or processing a creative thought and developing a play or crafting a poem. These precious moments are where creatives ideas begin. Thoughts are captured, collected and developed to form works of art - a slow but incredibly beautiful and intuitive process to witness and portray.
To view this piece on display and see more of my artwork come and visit no. 6 the mall opposite The Model in Sligo or explore my website for more irish inspired illustration.
I'm reconstructing muppet characters with an injection of madness throwing digital paint at pixels to create this monstrous piece of work with a unique and colourful style.
To see more illustration from me have a peek at the rest of my website or call into no 6 the mall and come say hi and see the only Sligo-based illustration studio.


text:
\section{Introduction}

The multiplicative ergodic theorem is a fundamental tool in the study of linear dynamical systems.
Given a Banach space $X$ write $\mathcal B(X)$ for the space of bounded linear operators on $X$.
Let $\Omega$ be a Lebesgue probability space.
Given an ergodic system $\sigma:\Omega\rightarrow\Omega$ and function $\mathcal L:\Omega\rightarrow\mathcal B(X)$, one may compose copies of $\mathcal L$ along orbits and investigate long term behaviour of any $x\in X$.
Such an $\mathcal L$ is referred to as a \textit{cocycle}, and is said to be \textit{forward-integrable} if $\log^+\Vert\mathcal L\Vert\in L^1(\Omega)$.
Iteratively set $\calL[0][\omega]=1_X:X\rightarrow X$, and $\calL[n][\omega]:=\mathcal L_{\sigma^{n-1}\omega}\circ\calL[n-1][\omega]$ for $n\in\mathbb N$.
The multiplicity property $\calL[m+n][\omega]=\calL[m][\sigma^n\omega]\circ\calL[n][\omega]$ is clear.
If $\mathcal L$ is required to be invertible then $\calL[n][\omega]=\mathcal L_{\sigma^n\omega}^{-1}\circ\cdots\circ\mathcal L_{\sigma^{-1}\omega}^{-1}$ for $n<0$ extends the cocycle to all $n\in\mathbb Z$.
If $\mathcal L^{-1}$ is forward-integrable then $\mathcal L$ is said to be \textit{backward-integrable}.
One may seek ways of describing $X$ in terms of long term behaviour of vectors under $\calL[n][\omega]$ as $n\rightarrow\infty$.
Given any normed space $V$ write $\mathbb S_V=\{x\in V:\Vert x\Vert=1\}$ and $\mathbb B_V=\{x\in V:\Vert x\Vert<1\}$.
Oseledets \cite{oseledets1968multiplicative} proved the following in 1965:
\begin{thm}
Let $\Omega$ be a Lebesgue probability space and $\sigma:\Omega\rightarrow\Omega$ be an invertible measure preserving transformation.
Let $\mathcal L:\Omega\rightarrow GL_d(\mathbb R)$ be forward and backward integrable, where $GL_d$ denotes the invertible $d-$dimensional matrices.
Then there are measurable numbers $\lambda_i(\omega),i\in\{1,\cdots,r_\omega\}$ and a direct sum decomposition of invariant measurable subspaces $\mathbb R^d=\bigoplus_iE_i(\omega)$ such that for $x\in\mathbb S_{E_i(\omega)}$,
$$
\lim_{n\rightarrow\pm\infty}\tfrac1n\log\Vert\calL[n][\omega]x\Vert=\pm\lambda_i(\omega).
$$
This limit converges uniformly in $x$.
\end{thm}
In a setting where all invertibility assumptions are dropped the conclusions are weaker.
The following flag decomposition is proven in the work of Raghunathan \cite{raghunathan1979proof}:
\begin{thm}
Let $\Omega$ be a Lebesgue probability space and $\sigma:\Omega\rightarrow\Omega$ be a (not necessarily invertible) measure preserving transformation.
Let $\mathcal L:\Omega\rightarrow\mat_d(\mathbb R)$ satisfy $\log^+\Vert\mathcal L\Vert\in L^1(\Omega)$ where $\mat_d$ denotes the $d$-dimensional matrices.
Then there are measurable numbers $\lambda_i(\omega),i\in\{1,\cdots,r_\omega\}$ and a flag of invariant measurable subspaces $\mathbb R^d=V_1(\omega)>V_2(\omega)>\cdots>V_{r_\omega}(\omega)$ such that for $x\in V_i(\omega)\setminus V_{i+1}(\omega)$,
$$
\lim_{n\rightarrow\infty}\tfrac1n\log\Vert\calL[n][\omega]x\Vert=\lambda_i(\omega).
$$
\end{thm}

These results were subsequently extended to more general settings: by Ruelle to compact operators on Hilbert spaces\cite{ruelle1979ergodic}, by Ma\~n\'e to compact operators on Banach spaces with additional continuity assumptions \cite{mane1983lyapounov} and by Thieullen to quasicompact operators \cite{thieullen1987fibres}.

The case where the underlying vector space is a separable Banach space was first presented by Lian and Lu \cite{lian2010lyapunov}.
Their monograph obtains the decomposition result assuming almost-everywhere injectivity of the cocycle in a separable Banach space.

The injectivity condition isn't necessary to obtain a decomposition: Froyland, Lloyd and Quas \cite{froyland2010semi} demonstrated that as long as $\sigma$ is invertible the space may still be written as a sum of fast and slow spaces.
That paper dealt with the finite dimensional case, but in \cite{gonzalez2011semi} Gonz\'alez-Tokman and Quas presented a generalisation to a separable Banach space.
However, there is some ambiguity to the statement and consequentially the proof - it is claimed that a measurable composition of $X$ exists without a precise discussion of what this means, which is crucial given that in the infinite dimensional setting no statement may be made of the slow spaces, as discussed in Horan's thesis \cite{horan2020spectral}.
Gonz\'alez-Tokman and Quas' subsequent shorter proof in the case where the dual $X^*=\mathcal B(X,\mathbb R)$ is separable used intuitive notions of volume growth that this work partially builds upon, but relied on less than fully rigorous references to cocycles whose domain varies depending on $\omega$.
Another example of this kind of emphasis on volume growth may be seen in the noninvertible result of Blumenthal in \cite{blumenthal2016vol} where a flag is obtained with no separability assumptions but with a much stronger uniform measurability condition for the cocycle.

Given measurable spaces $A$ and $B$ write $\mathcal M(A\rightarrow B)$ for the space of measurable functions from $A$ to $B$.
Let $\mathcal F$ be the Borel sigma algebra induced by the strong operator topology on $X$.
The space $\mathcal{SM}$ of strongly measurable functions consists of measurable maps with respect to this choice of sigma algebra:
$$
\mathcal{SM}\big(A\rightarrow\mathcal B(X))=\mathcal M(A\rightarrow(\mathcal B(X),\mathcal F)).
$$
Write $\mathcal G_kX=\{V\leq X:\dim V=k\}$ for the \textit{Grassmannian} of subspaces of dimension $k$.
Given $T\in\mathcal B(X)$ define the slowest growth of vectors in a given subspace under $T$:
$$
g(T,V)=g_T(V)=\inf_{x\in\mathbb S_V}\Vert Tx\Vert,\bern T=\sup_{V\in\mathcal G_kX}g_T(V).
$$
The $\rho_k$ are called Bernstein numbers in the work of Pietsch\cite{pietsch1974}, which gives an overview of similar kinds of statistics in Banach spaces.
Given a strongly measurable forward-integrable cocycle $\mathcal L$ ona Lebesgue probability space there are decreasing sequences $(\mu_i)_{i\in\mathbb N}$ and $\lambda_i$ of invariant functions
$$
\mu_k=\lim_{n\rightarrow\infty}\tfrac1n\rho_k\calL[n][\omega],\lambda_1=\mu_1,\lambda_{i+1}=\mu_{\inf\{t:\mu_t<\lambda_i\}}.
$$
While there are countably many $\mu_i$ there may only be finitely many $\lambda_i$s, referred to henceforth as the Lyapunov exponents.
The main result, an extension of the semi-inveritble result to separable Banach spaces, may now be stated:
\begin{thm}\label{mainres}
Let $(\Omega,\sigma,X,\mathcal L)$ be a strongly measurable forward-integrable random linear dynamical system with $(\Omega,\sigma)$ an ergodic invertible map on a Lebesgue probability space and having Lyapunov exponents $(\mu_i)_{i=1}^\infty$ and $(\lambda_i)_{i=1}^L$, where $1\leq L\leq\infty$.
Then for each $0\leq l<L$ there is a direct sum decomposition into  equivariant spaces $X=\big(\bigoplus_{i\leq l}E_i(\omega)\big)\oplus V_{l+1}(\omega)$ with the $E_l:\Omega\rightarrow\mathcal G_{m_l}X$ measurable, $m_l\in\mathbb N$ and having
$$
\tfrac1n\log\Vert\calL[n][\omega]\vert_{E_i(\omega)}\Vert,\tfrac1n\log\inf\{\Vert\calL[n][\omega]x\Vert:x\in\mathbb S_{E_i(\omega)}\}\rightarrow\lambda_i,
$$
and
$$
V_l(\omega)=\{x\in X:\lim_{n\rightarrow\infty}\tfrac1n\log\Vert\calL[n][\omega]x\Vert\leq\lambda_l\}.
$$
The projection $\Pi:X\rightarrow\bigoplus_{j<i}E_{j}$ parallel to $V_i$ is strongly measurable and \textit{tempered}, that is to say, $\lim_{n\rightarrow\infty}\tfrac1n\log\Vert\Pi_{\sigma^n\omega}\Vert=0$ almost surely.
There is a nontrivial decomposition, $L\geq2$, exactly when $\nu=\lim_{n\rightarrow\infty}\mu_n<\lambda_1$.
\end{thm}

The choice of construction method for the Lyapunov exponents is a fundamental aspect to each proof: in \cite{raghunathan1979proof} $\mu_i$ are written simultaneously using finite dimensional singular value decomposition, while by contrast in \cite{lian2010lyapunov} the first Lyapunov exponent is defined according to the asymptotic growth rate of $\Vert\calL[n][\omega]\Vert$ followed by the Lyapunov exponents being first described as the asymptotic growth rate for nonzero elements of the fast spaces in the statement of the thorem.
While singular value decomposition is no longer available in these contexts, some construction must be written down that may be viewed as reminiscent of a proof of the singular value decomposition of a transformation.
The work presented here explicitly relies on a notion of singular values for arbitrary elements of $\mathcal B(X)$.
Given a $T\in\mathcal B(X)$, sufficient conditions on these singular values for contracting fast growing regions of $\mathcal GX$ under the action $V\mapsto TV$ are derived.
A modified version of Kingman's subadditive theorem is established in reminiscent used to guarantee the asymptotic growth rates of the singular values of $\calL[n][\omega]$.

The results from past papers stated thus far have not required ergodicity.
Throughout the rest of this paper, ergodicity will be assumed, simplifying the classification of invariant functions, although all results as is typical may be formulated with this assumption dropped.
In this Banach space setting, the main theorem yields in a trichotomous classification of forward-integrable cocycles on separable Banach spaces: one of
\begin{itemize}
\item $\mathcal L$ fails to be quasicompact, with $\mu_i=\nu$ for all $i\in\mathbb N$ - no fast spaces may be detected
\item $\mathcal L$ is quasicompact, with finitely many finite dimensional fast spaces growing at rates $\lambda_1>\cdots>\lambda_L>\lambda_{L+1}=\nu$
\item $\mathcal L$ is quasicompact, with a countable sequence of finite dimensional fast spaces growing at rates $\lambda_1>\lambda_2>\cdots\rightarrow\nu$.
\end{itemize}
These three possibilities just correspond to situations where there are no, finitely many or countably many fast spaces.
The number $\nu$ is an alternative choice to the typical index of compactness $\kappa=\lim_{n\rightarrow\infty}\tfrac1n\Vert\calL[n][\omega]\Vert_c$ that proves simpler to work with in this proof.
In the appendix it is verified that these quantities are equal, so that the final result represents an extension of the result of Lian and Lu with the injectivity assumption dropped, avoiding measurability or domain concerns present in \cite{gonzalez2011semi} and \cite{gonzalez2015concise}.
In particular, the advantages of the current work are in its brevity, the preciseness of the conclusion and the use of an intuitive geometric perspective.

The author is indebted to Anthony Quas for his enthusiastic encouragement and guidance throughout.
\section{The Grassmannian}
When discussing subspaces of $X$, we may also consider the space of subspaces of $X$ they lie in:
\begin{defn}
Given vector spaces $U,V\leq X$ with $U\oplus V=X$, write $\Pi_{U\vert V}$ for the projection defined by $\Pi_{U\Vert V}(u+v)=u$ for any $u\in U$ and $v\in V$.
The Grassmannian is defined as the set
\begin{align*}
\mathcal GX=&\{V\text{ closed }\leq X:\text{ there exists a projection }X\xrightarrow\Pi V\text{ with }\Vert\Pi\Vert<\infty\}\\
=&\{V\text{ closed }\leq X:\text{ there exists }W\leq X\text{ with }\Vert\Pi_{V\Vert W}\Vert<\infty\}.
\end{align*}
$\mathcal GX$ may be metrised by any of a few equivalent choices of distance between spaces, such as the Hausdorff distance between unit spheres.
$\mathcal G_kX\subset\mathcal GX$ has already been defined.
Write $\mathcal G^kX=\{V\leq X:\codim V=k\}\subseteq\mathcal GX$.
\end{defn}
\begin{lem}\label{GXsep}
Suppose that $X$ is a Banach space.
Then
\begin{itemize}
\item
If $X$ is separable then $\mathcal G_kX$ is separable and complete.
\item
If $\Omega$ is a measurable space, $E\in\mathcal M(\Omega\rightarrow\mathcal G_kX)$ and $A\in\mathcal{SM}(\Omega\rightarrow\mathcal B(X))$ then the pointwise pushforward $\omega\mapsto A(\omega)E(\omega)$ is measurable.
\end{itemize}
\end{lem}
\begin{proof}
Completeness is established in section 2.1 of chapter IV of \cite{kato2013perturbation}.
The pushforward result is established in corollary B.13 of \cite{gonzalez2011semi}.
\end{proof}
\begin{lem}
If we take $B=\mathcal B(X)$ then we have the following characterisation of the space of strongly measurable functions:
$$
\mathcal{SM}\big(\Omega\rightarrow\mathcal B(X)\big)=\left\{\mathcal L\colon\Omega\rightarrow\mathcal B(X):\begin{array}{c}\text{for each }x\in X,\\(\omega\mapsto\mathcal L_\omega x)\in\mathcal M\big(\Omega\rightarrow X\big)\end{array}\right\}.
$$
\end{lem}
\begin{proof}
Lemma A.4 of \cite{gonzalez2011semi} checks the equivalence of these conditions.
\end{proof}
The following hold:
\begin{lem}\label{growth_ineq}
Let $T\in\mathcal B(X,Y)$ and $S\in\mathcal B(Y,Z)$.
\begin{itemize}
\item
$g_T(V)g_S(TV)\leq g_{S\circ T}(V)\leq\min\{\Vert T\vert_V\Vert g_S(TV),g_T(V)\Vert S\Vert\}$
\item
$\rho_k(T)\rho_k(S)\leq\rho_k(S\circ T)\leq\rho_k(T)\Vert S\Vert$
\end{itemize}
\end{lem}
We make use of the following straightforward construction:
\begin{lem}\label{lexstruct}
let $\Omega$ be a measurable space.
Suppose that $\{x_n\}_{n\in\mathbb N}\subseteq X$ for some measurable space $X$.
Then if we can write down a measurable set $S_n\subseteq\Omega$ for each $n\in\mathbb N$ such that $\bigcup_nS_n=\Omega$ then there is an associated measurable $N\colon\Omega\rightarrow\mathbb N$ given by $N_\omega=\inf\{n:\omega\in S_n\}$.
Further, the map $\omega\mapsto x_{N_\omega}$ is measurable.
\end{lem}
This principle is applied to check measurability in the context of spaces which are separable.
\section{Measurable statistics for linear operators}
\begin{lem}
Let $X$ be a normed space.
Then $g\colon\mathcal B(X)\times\mathcal G_kX\rightarrow[0,\infty)$ is continuous.
\end{lem}
\begin{proof}
Let $\epsilon>0$.
Here use the following metric inducing the product topology:
$$
d\big((S,V),(T,W)\big)=\Vert S-T\Vert+d(V,W).
$$
First note that $g$ is continuous at $(0,V)$ for any $V$ since $g(0,V)=0$ and if $d(V,W)+\Vert T-0\Vert<\epsilon$ then $g(T,W)<\epsilon$.
Otherwise, the projection onto the sphere $p_{\mathbb S}T=\tfrac1{\Vert T\Vert}T$ is continuous.
Let $\epsilon>0$.
Since $g(T,V)=\Vert T\Vert g(p_{\mathbb S}T,V)$ for nonzero $T$, it suffices to check $g$ is continuous on $\mathbb S_{\mathcal B(X)}\times\mathcal G_kX$.
Let
$$
(S,V),(T,W)\in\mathbb S_{\mathcal B(X)}\times\mathcal G_kX
$$
be such that
$$
d\big((S,V),(T,W)\big)=\Vert S-T\Vert+d(V,W)<\epsilon.
$$
Since $V$ and $W$ are finite dimensional their spheres are compact and we may choose an $x\in\mathbb S_V$ with $\Vert Sx\Vert=g(S,V)$.
Choose a $y\in\mathbb S_W$ with $\Vert x-y\Vert<\epsilon$.
Then
\begin{align*}
g(T,W)\leq&\Vert Ty\Vert\leq\Vert T-S\Vert+\Vert S\Vert\Vert(y-x)\Vert+\Vert Sx\Vert\\
<&\epsilon+\epsilon+g(S,V).
\end{align*}
By the same argument swapping $(S,V)$ and $(T,W)$ then, $g(S,V)<g(T,W)+2\epsilon$ and continuity (in fact, uniform continuity) on $\mathbb S_{\mathcal B(X)}\times\mathcal G_kX$ is clear.
Thus $g$ is continuous on $\mathcal B(X)\times\mathcal G_kX$ as required.
\end{proof}
\begin{cor}
The quantities $\rho_k:\mathcal B(X)\rightarrow\mathbb R$ are measurable functions.
\end{cor}
\begin{proof}
The $\rho_k$ may now be written as the supremum of the functions $\{T\mapsto g(T,V):V\in\mathcal G_kX\}$ which may be written as the supremum of a countable family since $\mathcal G_kX$ is separable.
\end{proof}
\begin{lem}{Grassmannian contraction estimates:}\label{pushclose}
Let $T\in\mathcal B(X)$ and $\Theta\in\left(\bern[k+1]T,\bern T\right)$.
Suppose that $V,W\in\mathcal G_kX$ are choices of fast growing spaces:
$$
g_T(V),g_T(W)>\Theta.
$$
Then
$$
d(TV,TW)<\frac{2}{1-\frac{\bern[k+1]T}\Theta}\frac{\bern[k+1]T}\Theta.
$$
In particular, if $\Theta>2\bern[k+1]T$ then
$$
d(TV,TW)<4\frac{\bern[k+1]T}\Theta.
$$
\end{lem}
\begin{proof}
If $V=W$ there is nothing to prove so assume otherwise.
Let $a\in V\setminus W$.
Then $W\oplus\spn\{a\}\in\mathcal G_{k+1}X$ so that there is some $a-b\in W\oplus\spn\{a\}$ with $b\in W$ and $g_T(a-b)\leq \bern[k+1]T$.
Set $c=a-b$.
By fastness $\Vert a\Vert\leq\frac{\Vert Ta\Vert}\Theta$ and $\Vert b\Vert\leq\frac{\Vert Tb\Vert}\Theta$.

\begin{align*}
\Vert Tc\Vert\leq\Vert c\Vert \bern[k+1]T&\leq(\Vert a\Vert+\Vert b\Vert)\bern[k+1]T\\
                                      &\leq\frac{\Vert Ta\Vert+\Vert Tb\Vert}\Theta \bern[k+1]T\\
                                      &\leq \frac{2\Vert Ta\Vert+\Vert Tc\Vert}\Theta \bern[k+1]T,
\end{align*}
So that
$$
\Vert Tc\Vert<\frac{2}{1-\frac{\bern[k+1]T}\Theta }\frac{\bern[k+1]T}\Theta \Vert Ta\Vert.
$$
Then since $a$ isn't in the kernel then $d(\frac{Ta}{\Vert Ta\Vert},TW)\leq\frac{\Vert Tc\Vert}{\Vert Ta\Vert}<\frac{2}{1-\frac{\bern[k+1]T}\Theta}\frac{\bern[k+1]T}\Theta $.
Since the choice of $a$ was arbitrary,
$$
d(TV,TW)<\frac{2}{1-\frac{\bern[k+1]T}\Theta }\frac{\bern[k+1]T}\Theta.
$$
\end{proof}
\begin{lem}
Let $k\in\mathbb N$.
Then there is a measurable $U:\mathcal G_kX\rightarrow\mathcal G_{k-1}X$ such that $U(V)<V$.
\end{lem}
\begin{proof}
Let $\{U_1,U_2,\cdots\}$ be dense in $\mathcal G_{k-1}X$.
Suppose $V\in\mathcal G_kX$.
Let $s_0(V)=1$ and
$$
s_i(V)=\inf\{s:\sup_{v\in\mathbb S_{U_s}}d(v,V)<2^{-i}\text{ and }d(U_s,U_{s_{i-1}(V)})<2^{1-i}\}\text{ for }i>0.
$$
In other words, $U_{s_i(V)}$ is the first member of the dense collection to fall in the nonempty open set
$$
B_{2^{1-i}}(U_{s_{i-1}})\cap\bigcup_{W\in\mathcal G_{k-1}X:W<V}B_{2^{-i}}(W).
$$
Then $U_{s_i}$ is Cauchy and so the pointwise limit of measurable functions $U_{s_i(V)}$ is convergent to some measurable $U(V)\in\mathcal G_{k-1}X$.
Further, $\sup_{v\in\mathbb S_{U(V)}}d(v,V)\leq 2^{-i}$ for any $i$, so $U(V)<V$.
\end{proof}
\begin{lem}
Let $k\in\mathbb N$.
Let $U\in\mathcal M(\mathcal G_kX\rightarrow\mathcal G_{k-1}X)$ with $U(V)<V$ as guaranteed by the previous lemma.
Then there is a measurable $b:\mathcal G_kX\rightarrow\mathbb S_X$ such that $U\oplus\spn\{b\}$ is the identity on $\mathcal  G_kX$.
\end{lem}
\begin{proof}
Let $\{x_1,x_2,\cdots\}$ be dense in $\mathbb S_X$.
Let $t_0(V)=1$ and
$$
t_i(V)=\inf\{t:d(U(V)\oplus\spn\{x_t\},V)<2^{-t}\text{ and }\Vert x_t-x_{t_{i-1}}\Vert<2^{1-i}\}\text{ for }i>0.
$$
Again we get that $x_{t_i}$ converges pointwise to some measurable $b$ with the desired properties.
\end{proof}
Applying this result recursively we obtain the following:
\begin{lem}\label{measbase}
Let $X$ be a separable Banach space.
Then for every $k>0$ there is a measurable map $b:\mathcal G_kX\rightarrow\mathbb S_X^k$ such that for each $V\in\mathcal G_kX$,
$$
V=\spn\{b_i(V)\}_{i=1}^k.
$$
\end{lem}
\begin{lem}\label{Tsmallgrow}
Let $V\in\mathcal G^kX$ with bounded projection $\Pi:X\rightarrow V$.
Let $T\in\mathcal B(X)$.
Then $\rho_{k+l}(T)\leq\rho_l(T\circ\Pi)$.
\end{lem}
\begin{proof}
\begin{align*}
\rho_{k+l}(T)=&\sup_{U\in\mathcal G_{k+l}X}g_T(U)\leq\sup_{U\in\mathcal G_{k+l}X}g_T(U\cap V)\\
=&\sup_{U\in\mathcal G_{\leq k+l,\geq l}V}g_T(U)=\sup_{U\in\mathcal G_lV}g_{T|_V}(U)\\
=&\sup_{U\in\mathcal G_lV} g_{T\circ\Pi}(U)\leq\rho_l(T\circ\Pi).
\end{align*}
\end{proof}
\begin{lem}\label{Tbiggrow}
Let $X=E\oplus V=E'\oplus V'$ with corresponding projections $\Pi:X\rightarrow V\in\mathcal G^kX$ and $\Pi':X\rightarrow V'\in\mathcal G^kX$.
Let $T\in\mathcal B(X)$ such that $\Pi'\circ T=T\circ\Pi$, and suppose further that $g_T(E)>\Vert T|_V\Vert$.
Then $\rho_l(T\circ\Pi)\leq4\Vert\Pi\Vert\Pi'\Vert\rho_{l+k}(T)$.
\end{lem}
\begin{proof}
Make use of the inequality $\rho_l(T\circ\Pi)\leq\Vert\Pi\Vert\rho_l(T|_V)$.
Let $U\in\mathcal G_lV$ with $g_{T\circ\Pi}(U)>e^{-\epsilon}\rho_l(\Pi\circ T|_V)=e^{-\epsilon}\rho_l(T|_V)$.
Let $x\in\mathbb S_{U\oplus E}$.
Observe that $\Vert\Pi'Tx\Vert\leq\Vert\Pi'\Vert\Vert Tx\Vert$ and similar for $1-\Pi'$ so that
\begin{align*}
\Vert Tx\Vert\geq&\max\{\tfrac{\Vert\Pi'T x\Vert}{\Vert\Pi'\Vert},\tfrac{\Vert(1-\Pi')Tx\Vert}{\Vert1-\Pi'\Vert}\}\geq\max\{\tfrac{g_{T\Pi}(U)\Vert\Pi x\Vert}{\Vert\Pi'},\tfrac{g_T(E)\Vert(1-\Pi)x\Vert}{\Vert1-\Pi'\Vert}\}\\
\geq&\tfrac1{1+\Vert\Pi'\Vert}\max\{g_{T\Pi}(U)\Vert\Pi x\Vert,g_T(E)\Vert1-\Pi x\Vert\}\\
\geq&\tfrac{e^{-\epsilon}\rho_l(T|_V)}{1+\Vert\Pi'\Vert}\max\{\Vert\Pi x\Vert,\Vert(1-\Pi)x\Vert\}\geq\frac{e^{-\epsilon}\rho_l(T|_V)}{4\Vert\Pi'\Vert}\geq\tfrac{e^{-\epsilon}\rho_l(T\circ\Pi)}{4\Vert\Pi'\Vert\Vert\Pi\Vert}.
\end{align*}
We may then conclude that $g_T(U)\geq\tfrac{e^{-\epsilon}\rho_l(T\circ\Pi)}{4\Vert\Pi'\Vert\Vert\Pi\Vert}$ with $\epsilon>0$ arbitrary, so that $\rho_l(T\circ\Pi)\leq4\rho_{l+k}(T\circ\Pi)\Vert\Pi'\Vert\Vert\Pi\Vert$ as required.
\end{proof}
\section{A balanced subadditive ergodic theorem}
Consider a decomposition of the interval $[a,b)=[a,c)\sqcup[c,b)\subset\mathbb Z$.
In discrete time, where these intervals represent different intervals in which dynamics may occur.
Given a cocycle $\mathcal L:\Omega\rightarrow\mathcal B(X)$, for fixed $\omega$ and each $b\geq a\in\mathbb Z$ we may define the map $\Lab(\omega):=\calL[b-a][\sigma^a\omega]$, which may be thought of as the evolution rule for $X$ from time $a$ to time $b$.
Under this notation it is easy to see that for any $a<c<b\in Z$, $\Lab[c][b]\circ\Lab[a][b]=\Lab$.
The inequality $\Vert\Lab\Vert\leq\Vert\Lab[a][c]\Vert\Vert\Lab[c][b]\Vert$ holds.
Taking $\log$ of each side, one obtains the following triangle inequality-like bound on the growth of points in $X$ from time $a$ to $b$:
$$
f_{a\rightarrow b}(\omega):=\log\Vert\Lab\Vert\leq\log\Vert\Lab[a][c]\Vert+\log\Vert\Lab[c][b]\Vert=f_{a\rightarrow c}(\omega)+f_{c\rightarrow b}(\omega).
$$
\begin{defn}
Let $\mathcal F=\{f_n\}_{n\in\mathbb N_0}\subseteq\mathcal M(\Omega\rightarrow\mathbb R)$ satisfy
$$
f_{m+n}(\omega)\leq f_m(\sigma^n\omega)+f_n(\omega).
$$
Then $\mathcal F$ is referred to as a subadditive family of measurable functions.
\end{defn}
The following view is useful:
\begin{defn}
A subadditive family $\{f_n\}_{n\in\mathbb N_0}$ generates a stationary subadditive process $\{f_{a\rightarrow b}:a<b,a,b\in\mathbb Z\}$ and vice versa via the relation
$$
f_{a\rightarrow b}:=f_{b-a}\circ\sigma^a.
$$
A subadditive process is a collection $\{f_{a\rightarrow b}\}_{a<b\in\mathbb Z}\subseteq\mathcal M(\Omega\rightarrow\mathbb R)$ such that for all $a<c<b\in\mathbb Z$,
$$
f_{a\rightarrow b}\leq f_{a\rightarrow c}+f_{c\rightarrow b},
$$
and the stationarity condition is
$$
f_{a+l\rightarrow b+l}=f_{a\rightarrow b}\circ\sigma^l.
$$
As such both notations may be interchanged as appropriate.
A similar formalism is outlined in \cite{krengel2011ergodic}.
\end{defn}

The Kingman theorem concerns subadditive families of functions, here a slight refinement when the underlying transformation is invertible will be required.
\begin{thm}[Kingman\cite{kingman1968ergodic}]\label{ing}Let $(\Omega,\sigma,\mathbb P)$ be an ergodic system and let $\{f_n\}_{n\in\mathbb N}\subset L^1\Omega$ be subadditive.
Then there is a constant $C\in[-\infty,\infty)$ such that
$$
\tfrac1nf_n(\omega)\rightarrow C=\lim_{n\rightarrow\infty}\tfrac1n\int f_n
$$
pointwise almost everywhere.
\end{thm}
In the case where $\sigma$ is invertible, one easily obtains the following useful corollary:
\begin{cor}
Let $(\Omega,\sigma,\mathbb P)$ be an invertible ergodic system on a Lebesgue probability space and let $\{f_n\}_{n\in\mathbb N}\subset L^1\Omega$ be subadditive.
Then
$$
\lim_{n\rightarrow\infty}\tfrac1nf_n(\sigma^{-n}\omega)=\lim_{n\rightarrow\infty}\tfrac1nf_n(\omega)
$$
pointwise almost everywhere.
\end{cor}
\begin{proof}
Simply set $g_n=f_n\circ\sigma^{-n}$.
Then $g_n$ is subadditive with respect to $\sigma^{-1}$:
\begin{align*}
g_{m+n}(\omega)=f_{m+n}(\sigma^{-(m+n)}\omega)\leq&f_m(\sigma^{-(m+n)}\omega)+f_n(\sigma^{m-(m+n)}\omega)\\
=& g_m(\sigma^{-n}\omega)+g_n(\omega),
\end{align*}
so that certainly there exists some $L$ such that $\tfrac1nf_n\circ\sigma^{-n}\rightarrow L$.
It remains to check that this limit and $\tfrac1nf_n(\omega)\rightarrow L'$ coincide.
Let $\epsilon>0$.
Then there exists an $N$ such that
$$
\mathbb P(\omega:\tfrac1nf_n(\omega)\in(L'-\epsilon,L'+\epsilon)>\tfrac12
$$
and
$$
\mathbb P(\omega:\tfrac1ng_n(\omega)=\tfrac1nf_n(\sigma^{-n}\omega)\in(L-\epsilon,L+\epsilon))>\tfrac12,
$$
so that the set
$$
\{\omega:\tfrac1nf_n(\sigma^n\omega)\in(L-\epsilon,L+\epsilon)\}\cap\{\omega:\tfrac1nf_n(\omega)\in(L'-\epsilon,L'+\epsilon)\}
$$
has positive measure, whence $|L-L'|<2\epsilon$.
$\epsilon$ was arbitrary though, so that $L=L'$.
\end{proof}
Kingman's original result may be modified in a few directions.
One may very rapidly obtain the following, which may be found in \cite{ferrando1995moving}:
\begin{lem}\label{easyking}
Let $\{f_n\}_{n\in\mathbb N}\subseteq L^1\Omega$ be a subadditive family of functions.
Then
$$
\tfrac1nf_{n\rightarrow2n}\rightarrow C=\lim_{n\rightarrow\infty}\tfrac1n\int f_n.
$$
\end{lem}
\begin{proof}
The lower bound is immediately clear, since
\begin{align*}
\tfrac1nf_{n\rightarrow2n}(\omega)\geq\tfrac1n(f_{0\rightarrow2n}-f_{0\rightarrow n})\rightarrow 2C-C=C\text{ as }n\rightarrow\infty.
\end{align*}
On the other hand, let $\epsilon>0$, then since $\tfrac1n\int f_n\rightarrow C$ there exists some $N$ such that for all $n\geq N$, $\tfrac1n\int f_n\leq C+\epsilon$.
Given some fixed $n\geq N$ and some $j<N$, set $j=\lfloor\tfrac nN\rfloor$ and break up the interval $[0,n)=[n,n+j)\sqcup[n+j,n+j+N),\cdots\sqcup[n+j+(t-1)N,n+j+\lfloor\tfrac nN\rfloor N)\sqcup[n+j+\lfloor\tfrac nN\rfloor N,2n)$.
Applying subadditivity in this manner
\begin{align*}
f_{n\rightarrow 2n}\leq&f_{n\rightarrow n+j}+\sum_{i=0}^t f_{n+j+iN\rightarrow n+j+(i+1)N}(\omega)+f_{n+j+tN\rightarrow n}\\
\leq& S_N|f_1|\circ\sigma^n(\omega)+\sum_{i=0}^tf_N\circ\sigma^{n+j+iN}(\omega)+S_N|f_1|\circ\sigma^{2n-N}(\omega).
\end{align*}
The above is valid for all $j\in[0,N)$, so averaging and dividing by $nN$ we obtain
\begin{align*}
\tfrac1nf_{n\rightarrow 2n}\leq&\tfrac1n(S_N|f_1|\circ\sigma^n+S_N|f_1|\circ\sigma^{2n-N})+\tfrac1{nN}S_nf_N\circ\sigma^n\\
\leq&2\epsilon+\tfrac 1{Nn}n(N(C+\epsilon))
\end{align*}
for sufficiently large $n$.
Since $\epsilon$ was arbitrary, the result is proven.
\end{proof}
A similar result for $f_{-n\rightarrow n}$ will be crucial to the main result.
The proof here uses the following lemma, which is a simplified version of result 3.9 in \cite{rudolph1990fundamentals}:
\begin{lem}[Backward Vitali]\label{bvit}
Let $(\Omega,\mathbb P,\sigma)$ be an invertible ergodic system on a Lebesgue probability space.
Suppose that there is a sequence of integer valued functions
$j_k\in\mathcal M(\Omega\rightarrow\mathbb N_0)$
such that $j_k(\omega)\rightarrow\infty$ as $k\rightarrow\infty$.
Then there is a measurable $A\subseteq\Omega$ and a measurable $j\in\mathcal M(A\rightarrow\mathbb N_0)$ such that writing $I_\omega=\{\sigma^i\omega:i\in\{-j_\omega,\cdots,j_\omega\}\}$,
the $I_\omega$ are disjoint and
$$
\mathbb P(\bigcup_{\omega\in A}I_\omega)>1-\epsilon.
$$
\end{lem}
\begin{thm}[Kingman's theorem for balanced intervals]\label{balancedkingman} let $(\Omega,\sigma,\mathbb P)$ be an ergodic system with invertible base, and let $(f_n)_{n\in\mathbb N}$ be  subadditive sequence of measurable functions on $\Omega$.
Then
$$
\tfrac1{2n}f_{2n}(\sigma^{-n}\omega)\rightarrow C=\lim_{n\rightarrow\infty}\tfrac1n\int f_n
$$
pointwise almost everywhere.
\end{thm}
\begin{proof}
Kingman's theorem immediately yields an upper bound: since
$$
\tfrac1{2n}f_{2n}(\sigma^{-n}\omega)\leq\tfrac1{2n}\big(f_n(\sigma^{-n}\omega)+f_n(\omega)\big)\rightarrow \tfrac12C+\tfrac12C
$$
it is clear that
$$
\bar C_\omega=\limsup_{n\rightarrow\infty}\tfrac1{2n}f_{2n}(\sigma^{-n}\omega)\leq C.
$$
If $C=-\infty$ then there is nothing more to show.
Otherwise,
\begin{align*}
\ubar C_{\sigma^{-1}\omega}=&\liminf_{n\rightarrow\infty}\tfrac1{2n}f_{2n}\circ\sigma^{-n}(\sigma^{-1}\omega)\\
\geq&\liminf_{n\rightarrow\infty}(\tfrac1{2n}(f_{2(n+1)}\circ\sigma^{-(n+1)}(\omega)-f_2\circ\sigma^{n-1}(\omega))\\
\geq&\liminf\tfrac1{2n}f_{2n}\circ\sigma^{-n}(\omega)-\limsup f_2\circ\sigma^{n-1}(\omega)\\
=&\ubar C_\omega-0,
\end{align*}
whence $\ubar C\leq\bar C$ is constant.
Fix $\epsilon>0$.
It is sufficient to show that $C-\epsilon\leq\Theta(\ubar C,\epsilon)$ where $\Theta(\ubar C,\epsilon)\rightarrow\ubar C$ as $\epsilon\rightarrow0$.
The set of $j$ such that $f_{-j\rightarrow j}$ is small is infinite, so denote the ordered elements
$$
\{j:f_{-j\rightarrow j}(\omega)<2j(\ubar C+\epsilon)\}=\{j_k(\omega):k\in\mathbb N\}.
$$
Since $f$ is integrable, it is uniformly integrable and so there exists a $\delta>0$ such that if $\mathbb P(A)<\delta$ then $\int_A f<\epsilon$.
Apply Backward Vitali to the $j_i$s and the $-j_i$s with parameter $\epsilon':=\min\{\epsilon,\delta\}$ to obtain an $A\subseteq\Omega$ and a $j:A\rightarrow\mathbb N$ according to theorem \ref{bvit} such that writing $B=\Omega\setminus\bigcup_{\omega\in A}I_\omega$, we have $\mathbb P(B)<\delta$ and so
$$
\int_{\Omega\setminus\bigcup_{\omega\in A}I_\omega}f_1^+<\epsilon.
$$
While every $j(\omega)=j_i(\omega)$, for some $i$ so that for every $\omega\in A$,
$$
f_{-j(\omega)\rightarrow j(\omega)}(\omega)\leq2j(\ubar C+\epsilon).
$$
Write $\Lambda_{\pm}=\{\sigma^{\pm j_i(\omega)}(\omega):\omega\in A\}$.
In addition we may define measurable maps $T_\pm\in\mathcal M(\Omega\rightarrow\mathbb N)$ by
\begin{gather*}
T_s(\omega)=\inf\{t\in\mathbb N_0:\sigma^t\omega\in\Lambda_-\},\\
T_e(\omega)=\inf\{t>T_s(\omega)\in\mathbb N_0:\sigma^{t}\omega\in\Lambda_+\}.
\end{gather*}
$T_s$ is the earliest nonnegative time such that $\sigma^{T_s(\omega)}\omega$ is the start of a period at growth rate guaranteed close to $\ubar C$, $T_e$ by contrast seeks the first time in the past which was the end of such an interval.
Each are almost surely finite since $\mathbb P(\Lambda_\pm)=\mathbb P(A)>0$.

The idea of the next step is that $[0,n)$ may be measurably broken up into intervals of slow growth and with small gaps inbetween.
Set $b_0=0$ and recursively iterate along the orbit for $i>0$: \begin{gather*}
a_i(\omega)=T_s(\sigma^{b_{i-1}(\omega)}\omega),\\
b_i(\omega)=T_e(\sigma^{b_{i-1}(\omega)}\omega).
\end{gather*}
For any $m\in\mathbb N_0$ we may set $t_m(\omega)=\max\{t:b_t(\omega)\leq m\}$.
Since $j$ and $T_\pm$ are measurable and the $t_m(\omega)\rightarrow\infty$ as $m\rightarrow\infty$, it is possible to pick an $N$ such that for all $n\geq N$, $\mathbb P(j\geq N)<\tfrac14\epsilon'$, $\mathbb P(t_n=0)<\tfrac14\epsilon'$ and $\mathbb P(T_\pm\geq N)<\tfrac14\epsilon'$.
Let $n\geq N$.
The orbit of $\omega$ is considered over the following intervals of times:
\begin{align*}
[0,n]=&[0,a_1(\omega))\cup[a_1(\omega),b_1(\omega))\cup\\
&[b_1(\omega),a_2(\omega))\cup[a_2(\omega),b_2(\omega)]\cup\\
&\cdots\\
&[b_{t_n(\omega)-1}(\omega),a_{t_n(\omega)}(\omega))\cup[a_{t_n(\omega)}(\omega),b_{t_n(\omega)}(\omega))\cup\\
&[b_{t_n(\omega)}(\omega),n).
\end{align*}
Time $a_i\rightarrow b_i$ is guaranteed to have low growth:
$$
f_{a_i(\omega)\rightarrow b_i(\omega)}(\omega)\leq(b_i(\omega)-a_i(\omega))(\ubar C_\omega+\epsilon),
$$
and in addition there is a good chance that the gaps between the $a_i$s and the $b_i$s won't be too large:
$$
\mathbb P(a_1(\omega)\geq N)=\mathbb P(T(\omega)\geq N)<\tfrac14\epsilon'.
$$
Since $T_-$ seeks the endpoints of the periods of guaranteed growth, it holds that either $T_-(\sigma^n\omega)=n-b_{t_n(\omega)}(\omega)$ or $t_n(\omega)=0$.
Therefore,
\begin{align*}
\mathbb P(n-b_{t_n(\omega)}(\omega)\geq N)\leq&\mathbb P(T_-(\sigma^n\omega)\geq N)+\mathbb P(t_n(\omega)=0)\\
<&\mathbb P(T_-\geq N)+\tfrac14\epsilon'<\tfrac12\epsilon'.
\end{align*}
Set
$$
\Lambda=\{a_1\geq N\text{ or }n-b_{t_n}(\omega)\geq N\},
$$
so that it is immediate from above that $\mathbb P(\Lambda)<\epsilon'$.
Given some measurable $g$ write $S_pg=\sum_{i=0}^{p-1}g\circ\sigma^i$.
We then repeatedly apply subadditivity to $f_{0\rightarrow n}$:
\begin{align*}
f_n(\omega)=&f_{0\rightarrow n}(\omega)\\
\leq&(f_{0\rightarrow a_1(\omega)}+f_{a_1(\omega)\rightarrow b_1(\omega)}+\cdots+f_{a_t(\omega)\rightarrow b_{t_{n\omega}}(\omega)}+f_{b_{t_{n\omega}}(\omega)\rightarrow n})(\omega)\\
\leq&\sum_{i=1}^{t_n(\omega)}f_{a_i(\omega)\rightarrow b_i(\omega)}(\omega)+\sum_{i=1}^{t_n(\omega)}f_{b_{i-1}(\omega)\rightarrow a_i(\omega)}(\omega)+f_{b_{t_n(\omega)}\rightarrow n}(\omega)\\
\leq&\sum_{i=1}^{t_n(\omega)}(b_i(\omega)-a_i(\omega))(\ubar C+\epsilon)+\\
&\sum_{i=a_1(\omega)}^{b_{t_n(\omega)}(\omega)}1_B(\sigma^i\omega)f_1\circ\sigma^i+f_{b_{t_n(\omega)}(\omega)\rightarrow n}(\omega).
\end{align*}
In the case where $t_n(\omega)=0$ the first two terms are empty sums.
Outside $\Lambda$ it is guaranteed that $[0,a_1(\omega))\subseteq[0,M)$ and $[b_{t_n(\omega)}(\omega),n)\subseteq [n-M,n)$.
Therefore
\begin{align*}
f_n(\omega)\leq&\sum_{i=0}^{n-1}(\ubar C+\epsilon)1_{B^c}(\sigma^i\omega)+\sum_{i=0}^{n-1}1_B(\sigma^i\omega)f_1(\sigma^i\omega)+\\
&\sum_{i=0}^M(|f_1|(\sigma^i\omega)+|\ubar C|+\epsilon)+\sum_{i=n-M}^{n-1}(|f_1|(\sigma^i\omega)+|\ubar C|+\epsilon)+\\
&1_\Lambda(\omega)S_n|f_1|(\omega)\\
=&(\ubar C+\epsilon)S_n1_{B^c}(\omega)+S_n(f_11_B)(\omega)+\\
&S_M|f_1|(\omega)+S_M|f_1\circ\sigma^{n-M}|(\omega)+1_\Lambda(\omega)S_n|f_1|(\omega).
\end{align*}
It remains to check that each of these terms are small enough to yield the result when we divide through by $n$ and integrate:
\begin{align*}
\tfrac1n\int(\ubar C+\epsilon)S_n1_{B^c}(\omega)+S_n(1_Bf_1)(\omega)=&\int_\Omega\big((\ubar C+\epsilon)1_{B^c}+f_11_B\big)\\
=&(\ubar C+\epsilon)\mathbb P(B^c)+\int_B|f_1|\\
\leq&\max\{(\ubar C+\epsilon)(1-\epsilon),\ubar C+\epsilon\}+\epsilon\\
\leq&(\ubar C+\epsilon)+\epsilon(|\ubar C|+\epsilon)+\epsilon,
\end{align*}
and
\begin{align*}
&\tfrac1n\int\big(S_M|f_1|(\omega)+S_M|f_1\circ\sigma^{n-M}|(\omega)+1_\Lambda(\omega)S_n|f_1|(\omega)\big)\\
=&\tfrac1n\int\big(M|f_1|+M|f_1|\big)+\tfrac1n\int_\Lambda S_n|f_1|\\
\leq&\frac{2M\epsilon}n+\tfrac1n\sum_{i=0}^{n-1}\int_{\sigma^i\Lambda}|f_1|\\
\leq&\frac{2M\epsilon}n+\epsilon.
\end{align*}
Putting these together we obtain
\begin{align*}
(C-\epsilon)\leq\tfrac1n\int_\Omega f_n\leq&(\ubar C+\epsilon)+\epsilon(|\ubar C|+\epsilon)+\epsilon+\frac{2M\epsilon}n+\epsilon\\
\rightarrow&\ubar C+(3+|\ubar C|)\epsilon\text{ as }n\rightarrow\infty.
\end{align*}

Letting $\epsilon\rightarrow0$ we obtain $C\leq\ubar C$ as required.
\end{proof}
Linearity of the cocycles yields subadditivity of these families, which allow us to apply Kingman.
\begin{defn}
Given a random linear dynamical system $\mathcal R=(\Omega,\sigma,X,\mathcal L)$, write
$$
\lambda_\omega(x)=\limsup_{n}\tfrac1n\log\Vert\mathcal L^{(n)}_\omega x\Vert\in[-\infty,\infty).
$$
Write $m_i$ for the multiplicity of $\lambda_i$, ie, $m_i$ is defined as the largest $m\in\mathbb N$ such that
$$
\mu_{m_1+\cdots+m_{i-1}+m}=\lambda_i.
$$
\end{defn}
\begin{lem}
The quantities $\lambda_i,\mu_i$ and $\nu$ are almost everywhere constants.
Further, $\lambda_2$ exists if and only if $\lambda_1>\nu$.
\end{lem}
\begin{proof}
The existence of the limits are guaranteed by applying Kingman to particular choices of subadditive sequences:
\begin{gather*}
f^k_n(\omega)=\log\bern(\calL[n][\omega]),\\
g_n(\omega)=\log\Vert\calL[n][\omega]\Vert_c.
\end{gather*}
The second statement is a trivial consequence of the definitions $\nu=\lim_{n\rightarrow\infty}\mu_n$ and $\lambda_2=\mu_{\inf\{t:\mu_t<\mu_1\}}$.
\end{proof}
\section{Decomposing a cocycle}
The tools obtained thus far are now used to decompose quasicompact cocycles.
\begin{lem}
Let $\mathcal R=(\Omega,\sigma,X,\mathcal L)$ be a strongly measurable random linear dynamical system with ergodic base on a separable Banach space.
Suppose that $\mathcal L$ satisfies the quasicompactness condition $\nu<\lambda_1$.
Then there is a unique measurable choice of fast space $E:\Omega\rightarrow\mathcal G_{m_1}X$ for which the following hold almost surely:
\begin{itemize}
\item equivariance: $\mathcal L_\omega E(\omega)=E(\sigma\omega)$
\item $\lambda(x)=\lambda_1$ for every $x\in E(\omega)\setminus\{0\}$
\item $\lim_{n\rightarrow\infty}\tfrac1n\log g(\calL[n][\omega],E(\omega))=\lambda_1.$
\end{itemize}
\end{lem}
\begin{proof}
Since $X$ is separable, by \cref{GXsep} $\mathcal G_{m_1}X$ is separable: choose some dense $\{E_i\}_{i\in\mathbb N}\subseteq\mathcal G_{m_1}X$.
Let $\epsilon\in(0,\tfrac15(\lambda_1-\lambda_2))$.
The sets
$$
A_i=\{\omega\in\Omega:g(\calL[2n],E_i)>e^{-\epsilon} \bern(\calL[2n]))\}
$$
are measurable and cover $\Omega$, since for fixed $\omega$, density of the $E_i$s and continuity of $g_{\calL[2n][\sigma^{-n}\omega]}$ means that we may find an $i$ with $g(\calL[2n][\sigma^{-n}\omega],E_i)$ as close to $\bern(\calL[2n][\sigma^{-n}\omega])$ as we please.
Then \cref{lexstruct} above provides measurable functions
\begin{gather*}
\iota^{(n)}(\omega)=\inf\{i\in\mathbb N:g(\calL[2n],E_i)>e^{-\epsilon}\bern (\calL[2n])\},\\
\tilde E^{(n)}(\omega):=E_{\iota^{(n)}\omega},
\end{gather*}
and the pushforward
\begin{gather*}
E^{(n)}(\omega)=\calL\tilde E^{(n)}(\omega)
\end{gather*}
is also then measurable by \cref{GXsep}.

First, we establish that this sequence is Cauchy, and therefore convergent, to a family of spaces $E\in\mathcal M(\Omega\rightarrow\mathcal G_{m_1}X)$.

For almost every $\omega\in\Omega$ the fastest $m_1$ dimensional growth rate
\begin{gather*}
\tfrac1n\log\bern(\calL)\rightarrow\lambda_1,\\
\tfrac1n\log\bern(\calL[n][\omega])\rightarrow\lambda_1,\\
\tfrac1{2n}\log\bern(\calL[2n])\rightarrow\lambda_1,\\
\end{gather*}
and
$$
\bern (\calL)\geq g_{\calL}(\tilde E^{(n)}(\omega))>e^{-\epsilon}\bern (\calL).
$$
Thus for each $\omega$ in this full measure set we may choose $M_\omega$ such that for $n\geq M_\omega$ we can usefully estimate growth under $\mathcal L$ in a few cases:
\begin{gather*}
\Vert\calL\Vert,\Vert\calL[n][\omega]\Vert\in(e^{n(\lambda_1-\epsilon)},e^{n(\lambda_1+\epsilon)}),\\
g(\calL[2n],\tilde E^{(n)}(\omega))\in(e^{2n(\lambda_1-\epsilon)},e^{2n(\lambda_1+\epsilon)})\\
\end{gather*}
and
$$
\Vert\mathcal L_{\sigma^{-(n+1)}\omega}\Vert<e^{n\epsilon}.
$$
Applying the inequalities in \cref{growth_ineq} then,
$$
g(\calL,\tilde E^{(n)}(\omega))\geq\frac{g(\calL[2n],\tilde E^{(n)}(\omega))}{\Vert\calL[n][\omega]\Vert}\geq e^{n(\lambda_1-3\epsilon)}
$$
and
$$
g(\calL[n][\omega],\calL\tilde E^{(n)}(\omega))=g(\calL[n][\omega],E^{(n)}(\omega))\geq\frac{g(\calL[2n],\tilde E^{(n)}(\omega))}{\Vert\calL\Vert}\geq e^{n(\lambda_1-3\epsilon)}.
$$
In addition, $\mathcal L_{\sigma^{-(n+1)}\omega}\tilde E^{(n+1)}(\omega)$ is also guaranteed to have fast growth under $\calL$:
\begin{align*}
g(\calL,\mathcal L_{\sigma^{-(n+1)}\omega}\tilde E^{(n+1)}(\omega)\geq&\frac{g(\calL[n+1][\sigma^{-(n+1)}\omega],\tilde E^{(n+1)}(\omega))}{\Vert\mathcal L_{\sigma^{-(n+1)}\omega}\Vert}\\
>&e^{(n+1)(\lambda_1-3\epsilon)}e^{-n\epsilon}\geq e^{n(\lambda_1-4\epsilon)}.
\end{align*}
$E^{(n)}(\omega)$ consists then of the image of vectors that were fast from time $-n$ to $0$, and will grow fast from time $0$ to $n$.
Then by \cref{pushclose} with $\Theta=e^{n(\lambda_1-4\epsilon)}$ we have
\begin{align*}
d(E^{(n)}(\omega),E^{(n+1)}(\omega))=&d\left(\calL\tilde E^{(n)}(\omega),\calL(\mathcal L_{\sigma^{-(n+1)}\omega}\tilde E^{(n+1)}(\omega))\right)\\
<&4\frac{e^{n(\lambda_2+\epsilon)}}{e^{n(\lambda_1-4\epsilon)}}<e^{-n(\lambda_1-\lambda_2-5\epsilon)}.
\end{align*}
Thus $E^{(n)}(\omega)$ is Cauchy and convergent since $\mathcal G_{m_1}X$ is complete, say to $E(\omega)$.

To prove equivariance, observe that for $n\geq\max\{M_\omega,M_{\sigma\omega}\}$, we find $\tilde E^{(n+1)}(\sigma\omega)$ is fast under $\calL$:
$$
g(\calL,\tilde E^{(n+1)}(\sigma\omega))\geq\frac{g(\calL[n+1][\sigma^{-n}\omega],E^{n+1}(\sigma\omega))}{\Vert\mathcal L_\omega\Vert}\geq e^{(n+1)(\lambda_1-\epsilon)}e^{-n\epsilon}>e^{n(\lambda_1-2\epsilon)}.
$$
Then, once again, by closeness of images of fast spaces,
$$
d(\calL[n][\sigma^{-n}\omega]\tilde E^{(n+1)}(\sigma\omega),E^{(n)}(\omega))<e^{-n(\lambda_1-\lambda_2-3\epsilon)},
$$
whence
\begin{align*}
d(E^{(n+1)}(\sigma\omega),\mathcal L_\omega E^{(n)}(\omega))=&d(\calL[n+1][\sigma^{-(n+1)}\omega]\tilde E^{(n+1)}(\sigma\omega),\mathcal L_\omega E^{(n)}(\omega))\\
\leq&e^{n\epsilon}d(\calL[n][\sigma^{-(n+1)}\omega]\tilde E^{(n+1)}(\sigma\omega),E^{(n)}(\omega))\\
<&e^{-n(\lambda_1-\lambda_2-4\epsilon)},
\end{align*}
so that $\mathcal L_\omega E(\omega)=\lim_{n\rightarrow\infty}\mathcal L_\omega E^{(n)}(\omega)=E(\sigma\omega)$.

To check that $E(\omega)$ is fast, choose $x\in\mathbb S_{E(\omega)}$.
Then since for $n\geq M_\omega$ we have $d(E^{(n)}(\omega),E(\omega))<e^{-n(\lambda_1-\lambda_2-\epsilon)}$, for each such $n$ we may choose an $x_n\in\mathbb S_{E^{(n)}(\omega)}$ with $\Vert x-x_n\Vert<e^{-n(\lambda_1-\lambda_2-\epsilon)}$.
Since $x_n\in E^{(n)}(\omega)$ and $g(\calL[n][\omega],E^{(n)}(\omega)\geq e^{n(\lambda_1-3\epsilon)}$ it then follows that,
\begin{align*}
\Vert\mathcal L^{(n)}_\omega x\Vert\geq&\Vert\mathcal L^{(n)}_\omega x_n\Vert-\Vert\mathcal L^{(n)}_\omega(x-x_n)\Vert\\
                                   \geq&e^{n(\lambda_1-3\epsilon)}-e^{-n(\lambda_1-\lambda_2-\epsilon)}\Vert\mathcal L^{(n)}_\omega\Vert\\
                               \implies&\Vert\mathcal L^{(n)}_\omega x\Vert\geq\tfrac12 e^{n(\lambda_1-\epsilon)}.
\end{align*}
Thus we may conclude that as well as being equivariant, $E(\omega)$ is fast for all sufficiently large $n$; since the choice of $x$ was arbitrary the growth is uniform:
$$
g(\calL[n][\omega],E(\omega))>e^{n(\lambda_1-\epsilon)}.
$$
On the other hand, for $n$ sufficiently large we also have
$$
g(\calL[n][\omega],E(\omega))\leq \bern (\calL[m][\omega])<e^{n(\lambda_1+\epsilon)},
$$
whence $\tfrac1n\log g(\calL[n][\omega],E(\omega))\rightarrow\lambda_1$.
Finally, we check uniqueness:  Suppose that $E(\omega)$ and $E'(\omega)$ are both equivariant and fast, so that for every $\omega$ there is some $N$ such that for $n\geq N$,
$$
g_{\calL[n][\omega]}(E(\omega)),g_{\calL[n][\omega]}(E'(\omega))>e^{n(\lambda_1-\epsilon)}.
$$
Define $\varphi\in\mathcal M(\Omega\rightarrow[0,1])$ by  $\varphi(\omega)=d(E(\omega),E'(\omega))$.
Applying lemma \ref{pushclose}, for almost every $\omega$ we have
\begin{align*}
\varphi(\sigma^n\omega)=&d(E(\sigma^n\omega),E'(\sigma^n\omega))=d(\calL[n][\omega]E(\omega),\calL[n][\omega]E'(\omega))\\
<&4\frac{\bern[k+1]\calL[n][\omega]}{e^{n(\lambda_1-\epsilon)}}\rightarrow0.
\end{align*}
$\varphi$ tends to zero along all orbits.
Therefore the sets $\{\varphi(\omega)>\epsilon\}$ all have measure zero, whence $\varphi$ vanishes almost everywhere and $E=E'$.
\end{proof}
The following lemma provides the top fast space for a general quasicompact cocycle:
\begin{lem}\label{topdecomp}
Let $\mathcal R=(\Omega,\sigma,X,\mathcal L)$ be a quasicompact, semi-invertible random linear dynamical system.
Then there exists a forward-equivariant decomposition $X=E(\omega)\oplus V(\omega)$ , where $V:\Omega\rightarrow\mathcal G^kX$ and the corresponding projection is a strongly measurable $\Pi:\Omega\rightarrow\mathcal B(X)$.
$V$ is a slow growing space: $\lim_{n\rightarrow\infty}\tfrac1n\log\Vert\calL[n][\omega]\vert_{V(\omega)}\Vert\rightarrow\lambda_2$ almost surely.
Finally, $\Pi_\omega$ is tempered.
\end{lem}
\begin{proof}
By \cref{measbase} we may choose a measurable family of bases
$$
(b_i)_{i=1}^k:\mathcal G_{m_1}X\rightarrow\mathbb S_X^k.
$$
Write $v_i(\omega)=b_i(E(\omega))$ which is itself then measurable.
Let $\{(q_{ij})_{i=1}^k\}_{j\in\mathbb N}$ be dense in $\mathbb R^k$.
Let $T_j\in\mathcal M(\Omega\times X\rightarrow X)$ be defined by
$$
T_j(\omega,x)=x-\sum_{i=1}^kq_{ij}v_i(\omega),
$$
so that for each $\omega\in\Omega$, the collection $\{T(\omega,\cdot)\}$ is dense in translations of $X$ by elements of $E(\omega)$.
Let $\epsilon<\tfrac12(\lambda_1-\lambda_2)$.
Define $\iota_n\in\mathcal M(X\setminus E(\omega)\rightarrow\mathbb N)$ for $n\in\mathbb N$ by
$$
\iota_n(\omega,x)=\inf\{j\in\mathbb N:\Vert\mathcal L^{(n)}_\omega T_jx\Vert\leq e^\epsilon \bern[k+1](\mathcal L_\omega^{(n)})\Vert T_jx\Vert\}.
$$
$\Pi^{(n)}\in\mathcal M(X\rightarrow X)$ may then be defined piecewise by
\begin{gather*}
\Pi^{(n)}_\omega(x)=
\begin{cases}
0\text{, if }x\in E(\omega),\\
T_{\iota_n(\omega,x)}x\text{ otherwise,}
\end{cases}\text{ and}\\
P^{(n)}_\omega(x)=x-\Pi^{(n)}_\omega(x).
\end{gather*}
To see that $\iota$ is finite on $X\setminus E(\omega)$, first note that the set
$$
\{T_jx:j\in\mathbb N\}
$$
is dense in the set $S=E(\omega)+x$.
The set
$$
U=\{y\in E(\omega)\oplus\spn\{x\}:\Vert\calL[n][\omega]y\Vert<e^{\tfrac12\epsilon}\bern[k+1](\calL[n][\omega])\Vert y\Vert\}
$$
is open and scale invariant - for every $\theta\neq 0$ we have $\theta U=U$, so
$$
U'=(E(\omega)+x)\cap U
$$
is open and nonempty in $S$, whence
$$
U'\cap\{T_jx\}_j\neq\emptyset,
$$
so that an $\iota_n(\omega,x)$ may be found in finite time.
To see that $\iota_n$ is measurable, note that
$$
\iota_n^{-1}\{1,\cdots,m\}=\bigcup_{i=1}^m\{(\omega,x):\Vert\calL[n][\omega]T_jx\Vert\leq e^\epsilon\bern[k+1](\calL[n][\omega])\Vert\calL[n][\omega]T_jx\Vert\}
$$
Let $x\in\mathbb S_X$.
Immediately from the definition, $P^{(n)}_\omega x\in E(\omega)$.
By convergence of the $\tfrac1n\log \bern $s, for all $n$ greater than or equal to some $N_\omega$,
\begin{gather*}
\Vert\mathcal L^{(n+1)}_\omega \Pi^{(n)}_\omega(x)\Vert\leq e^{(n+1)(\lambda_2+\epsilon)}\Vert\Pi^{(n)}_\omega x\Vert,\\
\Vert\mathcal L^{(n+1)}_\omega \Pi^{(n+1)}_\omega(x)\Vert\leq e^{(n+1)(\lambda_2+\epsilon)}\Vert\Pi^{(n)}_\omega x\Vert,\\
\Vert\mathcal L_{\sigma^n\omega}\Vert<e^{n\epsilon}\text{ and}\\
g_{\calL[n][\omega]}(E(\omega))\geq e^{n(\lambda_1-\epsilon)}.
\end{gather*}
In addition there is an easy bound independent of $x\in\mathbb S_X$ on $\Pi^{(n)}_\omega$:
\begin{align*}
\Vert\Pi^{(n)}_\omega x\Vert\leq&\Vert x\Vert+\Vert P^{(n)}_\omega x\Vert\\
\leq&1+e^{-n(\lambda_1-\epsilon)}\Vert\calL[n][\omega]P^{(N)}_\omega x\Vert\\
\leq&1+e^{-n(\lambda_1-\epsilon)}\big(\Vert\calL[n][\omega]x\Vert+\Vert\calL[n][\omega]\Pi^{(n)}_\omega x\Vert\big)\\
\leq&1+e^{-n(\lambda_1-\epsilon)}\big(e^{n(\lambda_1+\epsilon)}+e^{n(\lambda_2+\epsilon)}\Vert\Pi^{(n)}_\omega x\Vert\big),
\end{align*}
which rearranged yields
$$
\Vert\Pi^{(n)}_\omega x\Vert\leq\frac{1+e^{2n\epsilon}}{1-e^{-n(\lambda_1-\lambda_2-2\epsilon)}}\leq e^{3n\epsilon}.
$$
Consider differences between successive approximate slow components:
\begin{align*}
\Vert \Pi^{(n+1)}_\omega(x)-\Pi^{(n)}_\omega(x)\Vert=&\Vert P^{(n+1)}_\omega(x)-P^{(n)}_\omega(x)\Vert\\
\leq&\frac{\Vert\mathcal L^{(n+1)}_\omega(P^{(n+1)}_\omega x-P^{(n)}_\omega(x))\Vert}{g_{\calL[n][\omega]}(E(\omega))}\\
=&\frac{\Vert\mathcal L^{(n+1)}_\omega(\Pi^{(n+1)}_\omega x-\Pi^{(n)}_\omega(x))\Vert}{g_{\calL[n][\omega]}(E(\omega))}\\
\leq&e^{-(n+1)(\lambda_1-\epsilon)}\big(\Vert\calL[n+1][\omega] \Pi^{(n)}_\omega(x)\Vert+\Vert\calL[n+1][\omega] \Pi^{(n+1)}_\omega(x)\Vert\big)\\
\leq&e^{-(n+1)(\lambda_1-\epsilon)}\big(\Vert\mathcal L_{\sigma^n\omega}\Vert\Vert\calL[n][\omega] \Pi^{(n)}_\omega(x)\Vert+e^{n(\lambda_2+\epsilon)}\Vert\Pi^{(n+1)}_\omega(x)\Vert\big)\\
\leq&e^{-(n+1)(\lambda_1-\lambda_2-3\epsilon)}\big(e^{3n\epsilon}+e^{3(n+1)\epsilon}\big)\\
\leq&e^{-(n+1)(\lambda_1-\lambda_2-7\epsilon)},
\end{align*}
whence gaps between subsequent points decay exponentially, so that $(\Pi^{(n)}_\omega(x))_{n\in\mathbb N}$ forms a Cauchy and thus convergent sequence.
Set
$$
\Pi_\omega(x)=\lim_{n\rightarrow\infty}\Pi^{(n)}_\omega(x).
$$
Note that $\Pi_\omega\mathbb S_X$ is then bounded, since
\begin{align*}
\Vert\Pi_\omega x\Vert=&\lim_{m\rightarrow\infty}\Vert \Pi^{(m)}_\omega x\Vert\\
\leq&\Vert\Pi^{(N)}_\omega x\Vert+\lim_{m\rightarrow\infty}\Vert\Pi^{(N)}_\omega x-\Pi^{(m)}_\omega(x)\Vert\\
\leq&e^{3N\epsilon}+\frac{e^{-N(\lambda_1-\lambda_2-7\epsilon)}}{1-e^{-(\lambda_1-\lambda_2-7\epsilon)}},
\end{align*}
the final line being independent of choice of $x$.
Not only then is $\Pi^{(n)}_\omega x$ convergent, but we have the estimate
$$
\Vert\Pi^{(n)}_\omega x-\Pi_\omega x\Vert\leq\sum_{i=1}^\infty\Vert\Pi^{(n+i)}_\omega x-\Pi^{(n+i-1)}_\omega x\Vert<\frac{e^{-n(\lambda_1-\lambda_2-\epsilon)}}{1-e^{-\lambda_1-\lambda_2-\epsilon}}.
$$

$\Pi_\omega$ is linear: to see this let $b,c\in X$ and $t\in\mathbb R$, and set
$$
d:=\Pi_\omega^{(n)}(b+tc)-\Pi_\omega^{(n)}b-t\Pi_\omega^{(n)}c.
$$
Certainly $d\in E(\omega)$, since
\begin{align*}
d=&\Pi_\omega^{(n)}(b+tc)-(b+tc)+b-\Pi_\omega^{(n)}b+tc-t\Pi_\omega^{(n)}c\\
=&-P_\omega^{(n)}(b+tc)+P_\omega^{(n)}b+tP_\omega^{(n)}c\in E(\omega).
\end{align*}
Then applying the estimate for $\Pi^{(n)}_\omega$
\begin{align*}
\Vert d\Vert\leq&e^{-n(\lambda_1-\epsilon)}\Vert\calL[n][\omega]d\Vert\\
\leq&e^{-n(\lambda_1-\epsilon)}(\Vert\calL[n][\omega]\Pi^{(n)}_\omega(b+tc)\Vert+\Vert\calL[n][\omega]\Pi^{(n)}_\omega(b)\Vert+\Vert t\calL[n][\omega]\Pi^{(n)}_\omega c\Vert)\\
<&e^{-n(\lambda_1-\lambda_2-2\epsilon)}\big(\Vert\Pi^{(n)}_\omega(b+tc)\Vert+\Vert\Pi^{(n)}_\omega b\Vert+\Vert t\Pi^{(n)}_\omega c\Vert\big)\\
<&e^{-n(\lambda_1-\lambda_2-2\epsilon)}\big(\Vert\Pi_\omega(b+tc)\Vert+\Vert\Pi_\omega b\Vert+\Vert t\Pi_\omega c\Vert+\\
&(\Vert b+tc\Vert+\Vert b\Vert+\Vert tc\Vert)\frac{e^{-n(\lambda_1-\lambda_2-\epsilon)}}{1-e^{-\lambda_1-\lambda_2-\epsilon}}\big)\rightarrow0\text{ as }n\rightarrow\infty.
\end{align*}
Thus $d=0$ and $\Pi_\omega\in\mathcal B(X)$.
By construction $\Pi_\omega E(\omega)=0$.
On the other hand, $P_\omega X=E(\omega)$ since for any $x\in X$ and $n\in\mathbb N$ we have $x-\Pi^{(n)}_\omega x\in E(\omega)$.
$\Pi_\omega$ is then idempotent, since for any $x\in X$
$$
\Pi_\omega^2x-\Pi_\omega x=\Pi_\omega\circ P_\omega x\in\Pi_\omega E(\omega)=\{0\},
$$
and is thus a projection.
Set $V(\omega)=\Pi_\omega X$ so that $X=V(\omega)\oplus E(\omega)$.
For all $n\geq N_\omega$, and any $x\in V_\omega(x)$, because of the exponential rate of convergence there is a sequence of approximants
$$
\Vert x_n-x\Vert<C_\omega e^{-n(\lambda_1-\lambda_2-7\epsilon)}
$$
with
$$
\Vert\calL[n][\omega] x_n\Vert\in[0,e^{n(\lambda_2+\epsilon)}).
$$
Therefore,
$$
\Vert\calL[n][\omega]x\Vert<\Vert\calL\Vert\Vert x-x_n\Vert+\Vert\calL[n][\omega]x_n\Vert<e^{n(\lambda_2+8\epsilon)},
$$
whence
$$
\Vert\calL[n][\omega]\vert_{V(\omega)}\Vert<e^{n(\lambda_2+8\epsilon)}
$$
for $n\geq N_\omega$.
By the definition of $\lambda_2$, it is possible to choose $Y_n\in\mathcal G_{m_1+1}(\omega)$ with $\tfrac1ng_{\calL[n][\omega]}(Y_n)\rightarrow\lambda_2$, which means by dimension counting that there is always some $x_n\in\mathbb S_{V(\omega)\cap Y_n}$ with $\tfrac1n\Vert\calL[n][\omega]x_n\Vert\rightarrow\lambda_2$ and so
$$
\tfrac1n\log\Vert\calL[n][\omega]\vert_{V(\omega)}\Vert\rightarrow\lambda_2.
$$
Further, the map $\omega\mapsto\Pi_\omega$ is strongly measurable, since for each $x\in X$ the map $\omega\mapsto\Pi_\omega x$ is the limit of a sequence of measurable functions.

To see that $V(\omega)$ is equivariant it is sufficient to show that $P_{\sigma\omega}\circ\mathcal L_\omega x=0$ for all $x\in V(\omega)$.
If this were not the case, then $\mathcal L_\omega x$ would have a nonzero component in $E(\sigma\omega)$.
From this is would follow that $\lambda_\omega(x)=\lambda_1$ which would contradict the fact that $\Vert\calL[n][\omega]\vert_{V(\omega)}\Vert<e^{n(\lambda_2+8\epsilon)}$ for sufficiently large $n$.

As for temperedness of the projections:
Since $\Pi_\omega$ is bounded pointwise we may choose an $A\subseteq\Omega$ of positive measure on which $\Vert\Pi_\omega\Vert$ is at most some $M>0$.
Then define a new cocycle $\mathcal L'$ by
$$
\mathcal L_\omega'=\begin{cases}\mathcal L_\omega\circ\Pi_\omega\text{ if }\omega\in A,\\\mathcal L\text{ otherwise.}\end{cases}
$$
Then $\mathcal L_\omega'$ is forward-integrable since
$$
\int_\Omega\log^+\Vert\mathcal L_\omega'\Vert\leq \int_\Omega\log^+\Vert\mathcal L\Vert+\int_A\log^+M<\infty.
$$
Since $A$ has positive measure, there is almost surely an $n$ such that
$\calLp E(\omega)=0$,
and so
$$
\calLp X\subseteq V(\sigma^n\omega).
$$
We may then conclude that for each $x\in X$, $\lambda_\omega'(x)\leq\lambda_2$ and $\lambda_1'\leq\lambda_2$.
Applying lemma \ref{easyking} to the subadditive families $g_n=\log\Vert\calL[n][\omega]\Vert$ and $g_n'=\log\Vert\calLp\Vert$ there is an $N_1$ such that for $n\geq N_1$,
$$
\Vert\mathcal L'_{n\rightarrow2n}\Vert<e^{n(\lambda_2+\epsilon)}\text{ and }\Vert\mathcal L_{n\rightarrow2n}\Vert>e^{n(\lambda_1-\epsilon)}.
$$
Clearly $\mathcal L'_{n\rightarrow2n}\neq\mathcal L_{n\rightarrow2n}$ since the latter has a greater norm.
Therefore there must be some $j$ such that
$$
\mathcal L'_{n\rightarrow2n}=\mathcal L'_{n\rightarrow j}\circ\Pi_{\sigma^j\omega}\circ\mathcal L'_{j\rightarrow2n}=\mathcal L_{n\rightarrow2n}\circ\Pi_{\sigma^n\omega}.
$$
In addition, there exists an $N_2$ such that for $n\geq N_2$,
$$
g_{\calL[n][\sigma^n\omega]}(E(\sigma^n\omega))\in(e^{n(\lambda_1-\epsilon)},e^{n(\lambda_1+\epsilon)}).
$$
As a final condition, there exists some $N_3\in\mathbb N$ such that for all $n\geq N_3$, $g_{\calL[n][\omega]}(E(\omega))\in(e^{n(\lambda_1-\epsilon)},e^{n(\lambda_1+\epsilon)})$.
Let $x\in\mathbb S_X$.
Putting these together, for all $n\geq\max\{N_1,N_2,N_3,\tfrac1\epsilon\}$,
\begin{align*}
\Vert P_{\sigma^n\omega}x\Vert\leq&e^{-n(\lambda_1-\epsilon)}\Vert\calL[n][\sigma^n\omega]P_{\sigma^n\omega}x\Vert\\
\leq&e^{-n(\lambda_1-\epsilon)}\big(\Vert\calL[n][\sigma^n\omega]x\Vert+\Vert\calL[n][\sigma^n\omega]\Pi_{\sigma^n\omega}x\Vert\big)\\
\leq&e^{-n(\lambda_1-\epsilon)}\big(e^{n(\lambda_1+\epsilon)}\Vert x\Vert+\Vert\mathcal L_{n\rightarrow 2n\omega}'x\Vert\big)\\
\leq& e^{-n(\lambda_1-\epsilon)}\big(e^{n(\lambda_1+\epsilon)}+\Vert{\mathcal L'}_{n\rightarrow2n}\Vert\big)\Vert x\Vert\\
\leq& e^{2n\epsilon}+e^{-n(\lambda_1-\lambda_2-2\epsilon)}\leq e^{3n\epsilon}.
\end{align*}
$\epsilon$ was arbitrary and the norms of $\Pi$ and $P$ differ by at most $1$ so $\tfrac1n\log\Vert P_{\sigma^n\omega}\Vert,\tfrac1n\log\Vert\Pi_{\sigma^n\omega}\Vert\rightarrow0$ as required.
\end{proof}
\begin{cor}
$V(\omega)=\tilde V(\omega)=\{x\in X:\lambda_\omega(x)\leq\lambda_2\}$.
\end{cor}
\begin{proof}
$\tfrac1n\log\Vert\calL[n][\omega]\vert_V(\omega)\Vert\rightarrow0$ establishes the fact that $V(\omega)\subseteq\tilde V(\omega)$.
Conversely, any $x\in\tilde V(\omega)\setminus V(\omega)$ would have $P_\omega(x)\neq 0$ so that $\lambda_\omega(x)=\lambda_1$, contradicting the definition of $\tilde V$.
\end{proof}
\section{Proof of main result}
Finally, we may conclude with the main result, a well behaved decomposition of the space acted on by a random linear dynamical system:
\begin{proof}[Proof of thorem \ref{mainres}]
The decomposition is obtained inductively.
At each stage it is shown that if $\lambda_{i+1}$ exists, there exists an equivariant decomposition
$$
X=E_{\leq i}(\omega)\oplus V_{i+1}(\omega)
$$
with $E_{\leq i}\in\mathcal{SM}(\omega\rightarrow\mathcal G_{M_i}X)$ and bounded projections
\begin{gather*}
\Pi_{i+1\omega}:X\rightarrow V_{i+1}(\omega)\text{ and}\\
P_{i\omega}:X\rightarrow E_{\leq i}(\omega).
\end{gather*}
The existence of the top fast space has already been established - here denote this $E_{<2}(\omega)\oplus V_2(\omega)$ with measurable projections $\Pi_{2\omega}$ and $P_{2\omega}$.

Suppose that the statement is true up to $i=l-1$.
If $\lambda_l=\nu$ then we are done, so suppose otherwise - that there exists $\lambda_{l+1}\geq\nu$.
The projection $\Pi_{l\omega}$ is pointwise bounded, so that there exists some $M>0$ such that $A=\{\Vert\Pi_{l\omega}\Vert<M\}$ has positive measure.
$$
\mathcal L_\omega'=\begin{cases}\mathcal L_\omega\circ\Pi_{l\omega},\omega\in A,\\ \mathcal L_\omega\text{ otherwise.}\end{cases}.
$$
As before $\mathcal L'$ is forward integrable.
Write $\lambda',\mu_i',\lambda_i',\nu',E_l\oplus V',M_i',P'$ and $\Pi'$ for the asymptotic growth rates, Lyapunov exponents, decomposition, fast space multiplicities, fast projection and slow projection with $\mathcal L'$.
There almost surely exists an $N\in\mathbb N$ such that for each $n\geq N$, the followng hold:
\begin{itemize}
\item $\calLp=\calL[n][\omega]\circ\Pi_{l\omega}=\Pi_{l\sigma^n\omega}\circ\calL[n][\omega]$.
\item
$\Vert\Pi_{\sigma^n\omega}\Vert<e^{n\epsilon}$
\item
$g_{\calL[n][\omega]}(E(\omega))>\Vert\calL[n][\omega]|_{V_{l+1}(\omega)}\Vert.$
\end{itemize}
Therefore applying lemma\ref{Tsmallgrow} and \ref{Tbiggrow} to $\calLp=\calL[n][\omega]\circ\Pi_\omega$, for each $k$ the following holds:
$$
\rho_{k+M_l}(\calL[n][\omega])\leq\rho_k(\calLp)\leq4\Vert\Pi_{\sigma^n\omega}\Vert\Vert\Pi_\omega\Vert\rho_{k+M_l}(\calL[n][\omega])\leq 4e^{n\epsilon}\Vert\Pi_\omega\Vert\rho_{k+M_l}(\calL[n][\omega]),
$$
whence $\mu_{M_l+k}=\mu_k'$ and $\nu=\nu'$.
Further then, $m_k'=m_{l+k}$ and $\lambda+k'=\lambda_{k+l}$.
Set $V_{l+1}(\omega)=V_l(\omega)\cap V'(\omega)$ and $\Pi_{l+1\omega}=\Pi_{l\omega}\circ\Pi'_\omega$.
Write $E_{\leq l}(\omega)=E_{<l}(\omega)+E_l(\omega)$.
The equality $\mathcal L_\omega=\mathcal L_\omega'$ holds on $V_l(\omega)$, whence $\lambda_\omega=\lambda_\omega'$ on $V_l(\omega)$.
Since $A$ has positive measure, there is almost surely an $N$ such that for all $n\geq N$,
$\calLp E_{<l}(\omega)=0,~\calLp[n][\sigma^{-n}\omega]X\subseteq V_l(\omega)$,
and $\calLp X\subseteq V_l(\sigma^n\omega).$
We may then conclude that for each $x\in X$, $\lambda_\omega'(x)\leq\lambda_l$ and $\lambda_1'\leq\lambda_l$.
As such, for each $i<l$, have $E_i(\omega)\subseteq V'(\omega)$.
As for $E_l(\omega)$, by equivariance $E_l(\omega)=\calLp[n][\sigma^{-n}\omega] E(\sigma^{-n}\omega)\subseteq V_l(\omega)$.
Thus $X=E_{<l}(\omega)\oplus E_l(\omega)\oplus(V_l(\omega)\cap V'(\omega))=E_{<l}(\omega)\oplus E_l(\omega)\oplus V_{l+1}(\omega)$.

$\Pi_{l+1}$ is then also tempered:
$$
0\leq\tfrac1n\log\Vert\Pi_{l+1\sigma^n\omega}\Vert\leq\tfrac1n\log\Vert\Pi_{l\sigma^n\omega}\Vert+\tfrac1n\log\Vert\Pi_{\sigma^n\omega}'\Vert\rightarrow0.
$$
Let $\epsilon>0$.
There exists an $N$ such that for $n\geq N,~\Vert P_{\sigma^n\omega}'\Vert,\Vert\Pi_{\sigma^n\omega}'\Vert<e^{n\epsilon}$ and so for all $x\in X$ we have $\max\{\Vert P_{\sigma^n\omega}'\calL[n][\omega]x\Vert,\Vert\Pi_{\sigma^n\omega}'\calL[n][\omega]x\Vert\}\leq\Vert\calL[n][\omega]x\Vert$.

Rearranging this last inequality and letting $x\in E_{\leq l}(\omega)$,
\begin{align*}
\Vert\calL[n][\omega]x\Vert\geq&e^{-n\epsilon}\max\{\Vert P_{\sigma^n\omega}'\calL[n][\omega]x\Vert,\Vert\Pi_{\sigma^n\omega}'\calL[n][\omega]x\Vert\}\\
\geq&e^{-n\epsilon}\max\{\Vert\calL[n][\omega] P_\omega'x\Vert,\Vert\calL[n][\omega]\Pi_\omega'x\Vert\}\\
\geq&e^{-n\epsilon}\max\{e^{n(\lambda_l-\epsilon)}\Vert P_{\omega}'x\Vert,e^{n(\lambda_1'-\epsilon)}\Vert\Pi_\omega'x\Vert\}\\
\geq&e^{-n\epsilon}e^{n(\lambda_l-\epsilon)}\max\{\Vert\Pi_\omega'x\Vert,\Vert P_\omega'x\Vert\}\geq\tfrac12e^{n(\lambda_l-2\epsilon)}.
\end{align*}
For $n\geq\max\{N,\tfrac1\epsilon\log2\}$ it follows that $g_{\calL[n][\omega]}(E_{\leq l}(\omega))\geq e^{n(\lambda_l-3\epsilon)}$.
The characterisation
$$
V_{l+1}(\omega)=\{x\in X:\limsup_{n\rightarrow\infty}\tfrac1n\log\Vert\calL[n][\omega]x\Vert\leq\lambda_{l+1}\}
$$
holds.

\end{proof}

redpajama_set_name: RedPajamaArXiv


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Of what used to be us.
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As I decided not to see you again.
You know I don't have energy to waste for such feelings anyway. My memory should be more than enough.
I will just get back to conjugate myself with singular form. Once more.


Fast trains from Hereford to Bromsgrove take around 58 minutes, covering a distance of approximately 54 kilometres. There are frequent services on the rail route between Hereford and Bromsgrove.
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text:
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The Pittsburgh Climate Change Crankie is a manually powered paper panoramic illustration that tells the story of Pittsburgh's environmental evolution. The Crankie will help to start community conversations about Pittsburgh's industrial history, its environmental legacy, and the community's role in making improvements for the future through performances accompanied by hands-on art activities for all ages. The Crankie is illustrated by Katy DeMent and Laura Ramie.
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Note: like all village recipes, this simple dish relies on the freshest ingredients to give it its characteristic flavours. Harvest what you require just prior to cooking, don't be tempted to add or subtract from the ingredients, and resist the New Zealand urge to serve everything piping hot – this is room-temperature food, the flavour of which is greatly enhanced by 'sitting around' for a few hours (or even over-night) before being served with crusty bread.
Nick off the ends of each courgette and leave whole. Wash but do not scrape the potatoes. Heat half the olive oil in a saucepan. Add the courgettes and potatoes, stir to prevent browning, and cook for 5 minutes. Stir in the chopped dill, parsley, spring onions and spinach, add the salt and pepper, and cook for a minute or two. Add enough water to almost cover the vegetables. Simmer with the lid on the pot until the vegetables are tender. Remove pot from the heat and add the lemon juice and remaining olive oil. Leave the dish to cool to room temperature before serving.