Characterization of measurable sets in terms of open and compact sets I want to know if my way of proof this Proposition is true or not:
Characterization of measurable sets in terms of open and compact sets). Let E ⊂ R
d a bounded
set. Then, E is Lebesgue measurable if and only if for all ϵ>0 there exist an open set A and a compact set K with
K ⊆ E ⊆ A and m(A) − m(K)<ϵ
my proof:
if E is LM then $m_*(E)=m^*(E) $
$m_*(E)=sup(m(K), K compact, K\subset E) $
$m^*(E)=inf(m(A), K open, E\subset A) $
because both measures coincide and both are finite since E is bounded then there
exits a set $ A^0 open $ such that $ E\subset A^0 $ and 
$ m(A^0)<m^*(E)+\epsilon$
or
$ m(A^0)-\epsilon<m(E)$
also there exists a set $ K^0 compact$ such that 
$m_*(E)<m(K^0)+\epsilon$
or
$m(E)<m(K^0)+\epsilon$
so we get from the two inequalities that
$m(A^0)-\epsilon<m(K^0)+\epsilon$
$m(A^0)-m(K^0)<2\epsilon$
to prove the other direction
assume for all $\epsilon>0$ there exist
$A$ open and $K$ compact such that
$ K\subset E\subset A $
$m(A)-m(K)<\epsilon$
then
$m(A)<m(K)+\epsilon$
i take sup on right hand side and inf on the left hand side so that i get that
outer Lebsegue measure is less than inner measure for some arbitrary epsilon
then i let it epsilon goes to zero and conclude that both measure must coincide with each
other so that the set is Lebesgue measurable.
is this true
 A: I see two possible points of confusion.
First, what does it mean to "let $\epsilon$ go to zero"? In this part of the argument, the hypothesis is that for all $\epsilon>0$, there exist $K\subseteq E \subseteq A$ such that $m(A) - m(K) <\epsilon$. If you change "for all $\epsilon > 0$" to "for all $\epsilon \geq 0$", then the hypothesis is not satisfied by any $E$, so it's not surprising that you could conclude that $m^*(E) < m_*(E)$. 
So "let $\epsilon$ go to zero" must mean something slightly different. In fact, what you are doing is taking the inequality $m^*(E) < m_*(E) + \epsilon$, and taking an inf over all $\epsilon > 0$ to conclude that $m^*(E) \leq m_*(E)$ (and then applying the fact that we already know $m_*(E) \leq m^*(E)$ to conclude that they are equal). Notice that when you take an inf like this, $<$ gets converted $\leq$. It's worth thinking about why this is.
The fact that "$<$" becomes "$\leq$" also seems to be a point of confusion a step earlier in the argument (this is my second point). From $m(A) < m(K) + \epsilon$, you take a sup and an inf to conclude that $m_*(E) < m^*(E)+\epsilon$, but this isn't quite right -- either taking the sup or the inf on its own would be sufficient to convert the "$<$" to "$\leq$", so you can only conclude that $m_*(E) \leq m^*(E) + \epsilon$ in this step. But then you can still perform the final step in the same way as already disussed.
