Proof of consistency of Maximum Likelihood Estimator I would appreciate some help comprehending a logical step in the proof below about the consistency of MLE. It comes directly from Hogg, McKean, Craig, Introduction to Mathematical Statistics, 6th edition, Chapter 6, page 317.
So here goes:
Assume that $\hat{\theta_n}$ solves the estimating equation $\frac{\partial l(\theta)}{\partial \theta}=0$.
Also assume the following regularity conditions:

*

*the parameters identify the pdfs

*the pdfs  have common support for all $\theta$

*The point $\theta_0$ is the true parameter and is an interior point in $\Omega$
Then $\hat{\theta_n} \xrightarrow{P} \theta_0$
Proof
Let $\mathbf{X}=(x_1,x_2, \ldots, {x_n})$, the vector of observations.
Since $\theta_0$ is an interior point in $\Omega$ , $(\theta_0 -a, \theta_0 +a) \subset \Omega $ for some $a >0$. Define $S_n$ to be the event
$$S_n= \{ \mathbf {X} : l(\theta_0 ; \mathbf{X}) > l(\theta_0 -a ; \mathbf{X}) \}
 \cap \{ \mathbf{X}: l(\theta_0; \mathbf{X}) > l( \theta_0 +a  ;\mathbf{X}) \} $$
And this is where i have a problem. How can we evaluate the log-likelihood function at $\theta=\theta_0\pm a $ ? we only know that $(\theta_0 -a, \theta_0 +a) \subset \Omega $, we have not assumed that $\theta_0 -a\in\Omega$ or that $\theta_0 +a\in\Omega$. What if $\Omega=(\theta_0 -a, \theta_0 +a)$? Then the set $S_n$ is always empty(because $l(\theta_0 -a ; \mathbf{X})$ and $l(\theta_0 +a ; \mathbf{X})$ are undefined. ) and the proof breaks down for trivial reasons.
So that is my question. How is it justified to assume that $l(\theta_0 +a ; \mathbf{X})$ and $l(\theta_0 -a ; \mathbf{X})$ are defined?
Of course, the proof is not complete at this point but if I have this clarified, I can take it from there. Thank you in advance.
edit:
until now, I have seen this same proof in two books, namely Lehman Casella(2nd edition) page 447 and Hogg and Craig(6th edition) page 317. and in both books, the authors did not assume that $\theta-a$ and $\theta+a$ are inside the parameter space. And so, I am of the opinion that assuming $\theta-a\in\Omega$ and $\theta+a\in\Omega$ is unnecessary. What I do not understand is why these assumptions are unnecessary.

I am just a student but here an other prove that works if you all ready know by hypothesis that the MLE estimator is unbiased at least when the size sample goes to infinity.

*

*$\theta_n $ is the MLE estimator of an i.i.d sample $\left \{ X_1=x_1, X_2=x_2,\ldots,X_n=x_n \right \}$ of size $n$ and $\theta_0$ the parameter to estimate.
First let note that the sequence $(\theta_n)_n$ is a sequence of r.v. while $\theta_0$ is a parameter (a "constant").
So it is asking to us to prove: $$\forall \epsilon>0 \Rightarrow \lim_{n \to \infty }\mathbb{P}(|\theta_n - \theta_0| \geq \epsilon)=0$$


*In order to do that we will use the Markov Inequality with $r=2$ (the condition are verify by question hypothesis):
$$\forall \epsilon>0 \Rightarrow \mathbb{P}(|\theta_n - \theta_0| \geq \epsilon) \leq \frac{E(|\theta_n-\theta_0|^2)}{\epsilon^2}=\frac{E((\theta_n-\theta_0)^2)}{\epsilon^2}=\frac{E(\theta_n^2+\theta_0^2-2\theta_n \theta_0)}{\epsilon^2}=\frac{E(\theta_n^2)+\theta_0^2-2\theta_0E(\theta_n)}{\epsilon^2}$$


*As we know that $\theta_n$ is unbiased when $n$ goes to infinity it means that: $$\lim_{n \to \infty }E(\theta_n)-\theta_0=0 \Rightarrow \lim_{n \to \infty }E(\theta_n) = \theta_0 = E(\theta_0)$$
Hence $E(\theta_n)$ converges in law to $E(\theta_0)=\theta_0$. So because $f(x)=x^2$ is a continuous function:$$\lim_{n \to \infty }E(\theta_n^2)=E(\theta_0^2) = \theta_0^2$$
Q.E.D.
I hope this is correct.
