# Criterion for a set to be measurable

Okay. So I know that a set $E$ is called measurable if for any set $A$, we have $$m^\ast(A)=m^\ast(A\cap E)+m^\ast(A\cap E^c).$$ Recently, I came across a Lemma which says that

a set $A\subset E$ is called measurable if and only if $$m^\ast(A) + m^\ast(E\setminus A)=b-a$$ where $a$ and $b$ are the end points of the interval $E$.

I would like to know how to proof the above lemma. I have been trying for some time now with no results.
Thanks.

since $A\subset E$, $A\cap E=A$. so $m^\ast(A\cap E)=m^\ast(A)$. Also, $E\backslash A=E\cap A^c$. Thus $m^\ast(A) + m^\ast(E\setminus A)=b-a=m^\ast(E)$ implies that $$m^\ast(A\cap E) + m^\ast(E\cap A^c)=m^\ast(E).$$ Hence $A$ is measurable.

Conversely, suppose $A$ is measurable, then we have $$m^\ast(A\cap E) + m^\ast(E\cap A^c)=m^\ast(E).$$
since $A\subset E$, $A\cap E=A$ and $E\backslash A=E\cap A^c$, the result follows.

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 For this to make sense, $E$ must be an interval with endpoints $a$ and $b$, and you want $E \backslash A$, not $A \backslash E$. – Robert Israel Nov 27 '11 at 7:14 I had a typo earlier on, but I already fixed it. – AKM Nov 27 '11 at 7:41

In the lemma, we have $E\backslash A=E\cap A^c$, and $m^*(E)=b-a$. Does this help?
 Thanks for your answer. But if $A\subset E$, how is $E\backslash A=E\cap A^c$? – AKM Nov 27 '11 at 7:41 @AKM It should be easy to show $E\backslash A\subset E\cap A^c$ and $E\backslash A\supset E\cap A^c$ – Sid Raval Nov 27 '11 at 7:56 Oh okay...Thanks. I'll try and prove the lemma then. – AKM Nov 27 '11 at 8:21 I've some additions. Could please see if it's okay? Thanks. – AKM Nov 27 '11 at 8:33