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Let $(X,\mathcal{M},\mu)$ be a measure space such that $\mu(X)=1$, and let $\mu^{*}$ be the outer measure induced by $\mu$. Suppose $E\subset X$ satisfies $\mu^{*}(E)=1$. If $A,B\in \mathcal{M}$ and $A\cap E = B \cap E$, then $\mu(A)=\mu(B)$?

user206590
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2 Answers2

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Clearly $A$ and $B$ are $\mu^*$- measurable sets, therefore $\mu(A) = \mu^*(A)$ and $\mu(B) = \mu^*(B)$. Also note that $\mu^*(E^c) = 0$, because $\mu^*(E) = 1$.

$$\mu(A) = \mu^*(A) = \mu^*((A\cap E) \cup (A\cap E^c)) \leq \mu^*(A\cap E) + \mu^*(A\cap E^c) = \mu^*(A\cap E)$$

We know $\mu^*(A\cap E^c) = 0$ because $A\cap E^c\subset E^c$. Now we do the same reasoning for $B$.

$$\mu(B) = \mu^*(B) = \mu^*((B\cap E) \cup (B\cap E^c)) \leq \mu^*(B\cap E) + \mu^*(B\cap E^c) = \mu^*(B\cap E)$$

Finally, we know that $A\cap E = B\cap E$, therefore $\mu^*(A\cap E) = \mu^*(B\cap E) \implies$ $\implies\mu(A) = \mu(B)$.

Integral
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  • Comment left by Aspros189 as an answer: "Why $\mu^*(E^c)=0$? $E$ is not necessarily measurable." –  Jan 29 '15 at 19:56
  • This is outer measure, anything is "outer measurable". – Integral Jan 30 '15 at 01:56
  • How do you conclude $\mu^(E^c)=0$ from $\mu^(E)=1$, since additivity need not hold? –  Jan 30 '15 at 01:57
  • I think this is really a problem. I will try to fix, if possible. – Integral Jan 30 '15 at 02:15
  • To be honest, now I'm starting to think it's false and my answer is wrong. – Integral Jan 30 '15 at 02:24
  • So do I. Let $E$ be Vitali set, $A$ be the empty set, and $B$ a subset of $E^c$ of positive measure. –  Jan 30 '15 at 02:25
  • Should I delete my answer or let it there? – Integral Jan 30 '15 at 02:27
  • You could change it to the correct one... Here is a reference for the existence of a Vitali set of outer measure $1$. –  Jan 30 '15 at 02:28
  • Is there something special about the Vitali set here? Couldn't I just say that $E$ is not $X$ and not empty then work out the same arguments? Also, $E$ has full measure, to me is not obvious there is some subset of $E^c$ with positive measure. Isn't something we should prove before? – Integral Jan 30 '15 at 02:37
  • Yeah... don't have time right now, bookmarked this page. –  Jan 30 '15 at 02:40
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I just encountered this question and found an answer.

First, I proved every measurable set $F \subset E^c$ has measure zero:

$E \subset F^c \implies \mu^*(E) \leq \mu(F^c)$ and since $\mu^*(E) = 1$, $\mu(F^c) = 1$ and $\mu(F) = 0$

Now just write $A$ and $B$ as disjoint unions

$A = (A\cap B)\cup(A\setminus B)$

$B = (A\cap B)\cup(B\setminus A)$

and since $A\setminus B$ and $B\setminus A$ are in $E^c$, they have measure zero and $\mu(A)=\mu(B)$.

Bennie
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