3
$\begingroup$

Let $A\subseteq E\subseteq B\subset \mathbb R^d$ where $A,B$ are measurable and $m(A)=m(B)<\infty $. Show that $E$ is measurable.

In the correction they do like this: $$m^*(B\backslash E)\leq m^*(B\backslash A)=m(B)-m(A)=0.$$ Let $\varepsilon>0$ and Let $O\supset B$ an open set such that $$m^*(O\backslash B)<\varepsilon,$$ then $$m^*(O\backslash E)\leq m^*(O\backslash B)+m^*(B\backslash E)=m^*(O\backslash B)<\varepsilon$$ therefore $E$ is measurable.

Question: We just have shown that for all open set $O\supset E$ that contain $B$ (therefore $O$ depend of $B$ and $\varepsilon$ and not of $\varepsilon$ only), $$m^*(O\backslash E)\leq \varepsilon$$ not that for all open set $O\supset E$, $$m^*(O\backslash E)\leq \varepsilon$$

so we a priori can't conclude, no ?

P.S: $m$ is the Lebesgue measure and $m^*$ the exterior measure of Lebesgue.

$\endgroup$
  • $\begingroup$ Uh, $O \subsetneq E$. I am not sure what your question is. $\endgroup$ – IAmNoOne Jan 5 '15 at 11:25
  • $\begingroup$ I corrected it, thanks. I think is going to be more clear, isn't it ? $\endgroup$ – idm Jan 5 '15 at 11:27
2
$\begingroup$

This is because for any open set $O$ in the inclusion $E\subseteq O\subseteq B$, we would have $$O - E \subseteq B - E.$$

Since $m^*(B - E) = 0$, it follows that $m^*(O - E) = 0.$

This is true in general, that is,

If $m(Y) = 0$, then for any subset $X \subseteq Y$ the measure $m(X) = 0$.

$\endgroup$
  • $\begingroup$ nice :-) tks !! $\endgroup$ – idm Jan 5 '15 at 22:46
1
$\begingroup$

The proposed solution is assuming the following fact:

A set $E$ is measurable iff for every $\varepsilon > 0$, there exists an open sets $U$ such that $U \supseteq E$ and $m^{\star}(U \backslash E) < \varepsilon$.

I am not sure why this fact is being invoked. Whatever your official definition of measurability might be, it should be fairly straightforward to see that adding or subtracting a measure zero set to a measurable set results in a measurable set.

$\endgroup$
0
$\begingroup$

I would start how they start, which shows $m^*(B\setminus E)=0$. This implies $B\setminus E$ is measurable. Hence $E=(B^c\cup(B\setminus E))^c$ is measurable.

$\endgroup$
  • $\begingroup$ I know this solution, it's not what I'm looking for ! I want to understand why the can conclude whereas they show only that for all open $O\supset E$ that contain $B$, $m^*(O\backslash E)<\varepsilon$ and not that for all open $O\supset E, m^*(O\backslash E)<\varepsilon$ (I'm thing for exemple to the open set $E\subset O\subset B$. $\endgroup$ – idm Jan 5 '15 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.