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I have the following exercise I wish to solve:

Let $(X,S,\mu)$ be a measure space s.t $\mu(X)=1$.

Let $\mu^{*}$be defined on $X$ by:

$\forall E\subseteq X:\,\mu^{*}(E):=\inf\{\sum_{i=1}^{\infty}\mu(A_{i})\,\mid\, A_{i}\in S,E\subseteq\cup A_{i}\}$

Let $E\subseteq X$ s.t $\mu^{*}(E)=1$, prove that if $A,B\in S,A\cap E=B\cap E$ then $\mu(A)=\mu(B)$

What I tried is to write $$A=(A\cap E)\cup(A\cap E^{c}),B=(B\cap E)\cup(B\cap E^{c})$$ and then apply $\mu^{*}$on both of them.

Since $$A\cap E=B\cap E$$ I would have had $$\mu^{*}(A\cap E)=\mu^{*}(B\cap E)$$ and I hoped that $$\mu^{*}(E)=1\implies\mu^{*}(E^{c})=0$$ and from $\mu^{*}$being monotone that means that $$\mu^{*}(A\cap E^{c})=\mu^{*}(B\cap E^{c})=0$$

hence $\mu^{*}(A)=\mu^{*}(B)$ but $A,B\in S$ hence $$\mu^{*}(A)=\mu(A),\mu^{*}(B)=\mu(B)$$ and than I would be done.

But from this post is turns out that $\mu^{*}(E^{c})$ doesn't have to be $0$ and so my argument is not valid.

Can someone please suggets how to solve this problem ? what I tried was my only idea.

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Hint: Without loss of generality, assume that $A \subset B$ (otherwise replace $(A,B)$ by $(A, A \cup B)$. We need to show that $B - A$ has measure 0, and we know that it lies in the complement of $E$. Try to show that if $\mu ( B - A) > 0$, then together with $\mu^*(E) = 1$, it would imply that $\mu(X) > 1$. –  user27126 Dec 17 '12 at 11:15

1 Answer 1

up vote 2 down vote accepted

If $\mu(A)<\mu(B)$, then $\mu(B\backslash A)>0$. Hence $\mu\big(X\backslash (B\backslash A)\big)<1$. Now $E\subseteq (X\backslash (B\backslash A)\big)$ since $A$ and $B$ agree on $E$. Hence $\mu^*(E)<1$.

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Are you assuming $A\subseteq B$ ? I don't understand what to do if that is not the case –  Belgi Dec 17 '12 at 17:52
    
@Belgi No. $B=(B\backslash A)\cup(B\cap A)$ and hence $\mu(B)=\mu(B\backslash A)+\mu(B\cap A)$. Hence $\mu(B)-\mu(B\cap A)=\mu(B\backslash A)$. Now, $\mu(B\cap A)\leq \mu(A)$ , so $\mu(B)-\mu(A)\leq\mu(B\backslash A)$. –  Michael Greinecker Dec 17 '12 at 18:02
    
Can you please explain why $E\subseteq (X\backslash (B\backslash A)\big)$ ? –  Belgi Dec 17 '12 at 18:55
    
$A \cap E = B \cap E$ implies that $(B \backslash A )\cap E = \emptyset$, which implies that. –  user27126 Dec 17 '12 at 20:41
    
@Belgi Suppose $x\in E$, but $x\notin (X\backslash (B\backslash) A)$. Then $x\in B\backslash) A$. Hence $x\in B\cap E$ and $x\notin A\cap E$, contradicting the assumption that $B\cap E=A\cap E$. –  Michael Greinecker Dec 18 '12 at 8:28

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