Vtyjkyuk

Let $mathbbR^nsupsetI=(a,b)=(a_1,b_1)times...times(a_n,b_n)$ with $a,binmathbbR^n$ such that $a_ilt b_i
$$forall i$.
So that the outer measure is defined as:

$$
mu^*(A)=infsum_iinJV(I_i)mid Asubsetbigcup_iinJ(I_i)
$$

Where:
$I_i_iinJ$ is a cover of A.

I have already proven that every $I$ is a Lebesgue-measurable set. Now I want to prove that every open set is Lebesgue-measurable.

I have done this, is it correct?

Let A be an open set. For every $xin A$ with $x=(x_1,..,x_n)$ there exist an $r>0 $ such that $B(x;r)subset A$. If we take $r$ small enough, there exists $alpha > 0$ such that:

$$
B(x;r)subset I_x=(x_1-alpha,x_1+alpha)times...times(x_n-alpha,x_n+alpha)subset A
$$

Then define $A$ with:
$$
A=bigcup_xin A I_x
$$
A is union of Lebesgue-measurable sets so A is Lebesgue-measurable.

No, your argument does not work as is, because it is not true that an arbitrary union of measurable sets is measurable (if it were, every set would be measurable since points are measurable).

What you need to do is to write $A$ as a countable union of certain $I_n$. You can achieve this for instance by considering the set
$$
A_r=xin A: x_kinmathbb Q, k=1,ldots,n.
$$
For each $xin A_r$ there exists $I_x,alpha=(x_1-alpha,x_1+alpha)timesldotstimes(x_n-alpha,x_n+alpha) $ such that $I_x,alphasubset A$. Then, if we let $I_x,alpha=emptyset$ when $I_x,alphanotsubset A$,
$$
A=bigcup_xin A_rbigcup_alphainmathbb Q_+ I_x,alpha
$$
is a countable union of intervals, so measurable.