Question: This is actually from Zurich's book (chapter on Integration).

> 
> **Definition.** A set $E\subset \mathbb{R}$ *has measure zero* or *is of measure zero* (in the sense of Lebesgue) if for every number
> $\epsilon>0$ there exists a covering of the set $E$ by an at most
> countable system $\{I\_k\}$ of intervals, the sum of whose lengths
> $\sum \limits\_{k=1}|I\_k|$ is at most $\epsilon$.
> 
> 
> **Lemma.** *The union of a finite or countable number of sets of measure zero is a set of measure zero.*
> 
> 
> 

*Proof:* Suppose that $E=\cup\_{n=1}^{\infty}E\_n$ and each $E\_n$ has measure zero. Suppose $\epsilon>0$ be given. For each $n\in \mathbb{N}$ by definition we can find an at most countable system $\{I\_{n,k}\}\_{k\geq 1}$ of intervals such that $\sum \limits\_{k=1}^{\infty}|I\_{n,k}|<\dfrac{\epsilon}{2^n}$.

Then we see that $\{I\_{n,k}: n,k\geq 1\}$ be a countable system of intervals covering $E$.

How formally to show that total length of these $I\_{n,k}$ is less than $\epsilon$?

Because if we want to compute its total length then we need to deal with the double sum $\sum \limits\_{n,k\geq 1}$ or $\sum \limits\_{n\geq 1} \sum \limits\_{k\geq 1}
$ (btw I don't know the difference between them).

I would be very thankful if someone can show the formal proof of it please!

Answer: 
I don't think Zorich is super clear on this point. The closest he comes to an explanation is this:

> 
> The order of summation $\sum\_{n,k} |I^n\_k|$ on the indices $n$ and $k$ is of no importance, since the series converges to the same sum for any order of summation if it converges in even one ordering. [Zorich is referring to Proposition 4, page 270.] Such is the case here, since any partial terms of the sum are bounded above by $\varepsilon$.
> 
> 
> 

I would explain this this way. Let $a\_n = \sum\_{k} |I^n\_k| < \frac{\varepsilon}{2^n}$. Now let $A = \sum\_{j=1}^J |I^{n\_j}\_{k\_j}|$ be any partial sum of the series $\sum\_{n,k} |I^n\_k|$ arranged in an arbitrary order. If we let $N = \max\_j n\_j$, then $A \leq \sum\_{n = 1}^N a\_n \leq \sum\_{n = 1}^{+\infty} a\_n$.

Since any partial sum of the series $\sum\_{n,k} |I^n\_k|$ is $\leq \sum\_{n = 1}^{+\infty} a\_n$, we have
$$
\sum\_{n,k} |I^n\_k| \leq \sum\_{n = 1}^{+\infty} a\_n < \varepsilon.
$$

It's true there are theorems about double series that would make the original claim obvious, but you can get by without them.
The point here is that you don't actually need to compute it. What you do is for your first set, $E\_1$, choose a covering so that the total measure is less than $\frac \epsilon 2$. The second set you choose one whose total is less than $\frac \epsilon 4$, then $n'th$ one a cover whose total length is less than $\frac \epsilon {2^n}$. Now, taking the union and all you have is the simple geometric series $\frac \epsilon {2^n}$, which adds up to $\epsilon$

Edited to add complicated notation:

For Set $E\_1$ choose a covering $E\_{1,k}$ such that $\sum\_{k\geq 1}E\_{1,k}<\frac \epsilon 2$. Similarly, for each set $E\_n$ choose a covering so that $\sum\_{k\geq 1}E\_{n,k}<\frac \epsilon {2^n}$

Now, $\bigcup\_{n\geq 1}E\_n $is covered by $\bigcup\_{n\geq 1}\bigcup\_{k\geq 1}E\_{n,k}$, and taking the sum of the lengths, we have the union is less than

$$\sum\_{n\geq 1}\sum\_{k\geq 1}E\_{n,k}=\sum\_{n\geq 1}\left(\sum\_{k\geq 1}E\_{n,k}\right)=\sum\_{n\geq 1}\frac \epsilon {2^n}=\epsilon $$