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could any one help me to solve the problem?

$E\subseteq M\subseteq \mathbb {R}$, $M$ is measurable given that outer measure $m$ of $M<\infty$, we need to show $E$ is measurable iff $m(M)=m(E)+m(M\setminus E)$

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What definition of measurable set are you using? –  tomasz Aug 22 '12 at 1:44
    
$E\subseteq\mathbb {R}$ is measurable if for any $A\subseteq \mathbb R$, $m(A)=m(A\cap E)+ m(A\cap E^c)$ –  El Angel Exterminador Aug 22 '12 at 1:50
    
In this case, this problem seems to need nothing more than checking (for $E$) and applying (for $M$) the definition. What's stumping you? –  tomasz Aug 22 '12 at 1:53
    
where the use of $M$ is measurable? –  El Angel Exterminador Aug 22 '12 at 1:54
    
Hint: Via this definition, a set is measurable iff its complement is measurable, and iff its relative complement with a measurable set is measurable. –  Eugene Shvarts Aug 22 '12 at 1:55

1 Answer 1

Notice that $\Rightarrow$ is trivial, so you only really need to show the other one.

Because $M$ is measurable, we can assume that $A\subseteq M$, and because it is of finite measure, we can assume that so are all the involved sets.

We want to have $$\inf\{ m(U)\vert U\supseteq A\}=\inf\{ m(U)\vert U\supseteq A\cap E\}+\inf\{ m(V)\vert V\supseteq A\setminus E \}$$ which can be rewritten as $$\inf\{ m(U)\vert U\supseteq A\}=\inf\{ m(U)+m(V)\vert U\supseteq A\cap E,V\supseteq A\setminus E \}$$ and further $$\inf\{ m(U)\vert U\supseteq A\}=\inf\{ m(U\cup V)+m(U\cap V)\vert U\supseteq A\cap E,V\supseteq A\setminus E \}$$ But notice that by the assumption for each $n$, we have $U_n,V_n$ measurable such that $U_n\supseteq E$ and $V_n\supseteq M\setminus E$ and $m(U_n\cap V_n)<1/n$ (to show this, you need to use finiteness).

In particular, can assume that $(U\cap V)$ is of very small measure in the right hand side, yielding the equality. I will leave the technical details to you.

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