# 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

• It takes a lot of effort to read through another mathematician's proof line-by-line without any context, so you are unlikely to get any feedback to this question as stated. You will be more likely to get feedback if you ask a more precise question. For instance, is there a particular step in the proof that you're unsure about? Or are you perhaps unsure whether the result you're trying to prove is actually true? – tcamps Dec 26 '14 at 23:23
• Your proof is fine if your definition of measurability is indeed that $m_\ast (E)=m^\ast (E)$, with $m_\ast,m^\ast$ defined as in your answer (or if you have such a characterization). But observe that this characterization is in general false for unbounded $E$. – PhoemueX Dec 27 '14 at 8:56
• @tcamps the thing that i am not sure about it is the last part when i reach that m(A) is less that m(K) plus epsilon because when i take sup on the RHS and inf on the LHS i will get that the outer measure is strictly less than the inner measure plus epsilon then i let epsilon zero and i get that outer measure is strictly less than the inner measure which does not seem logic. – Saddam Hijazi Dec 27 '14 at 15:59
• @PhoemueX yes this characterization is just for bounded sets – Saddam Hijazi Dec 27 '14 at 16:00

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.