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.

  • $\begingroup$ 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

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 $m(X) = 0$.

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

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.


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.

  • $\begingroup$ 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

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