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So, this should be a little obvious, but not to me at the moment.

Why is $m(B = E - A) = m(E) - m(A)$?

Of course, all given sets are measurable.

Thanks.

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Please write the question in the body as well. –  Asaf Karagila May 14 '11 at 16:52
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2 Answers 2

up vote 3 down vote accepted

This is false. For example, let $m$ be counting measure on (the power set of) $\{1,2\}$, let $B=E=\{1\}$ and let $A=\{2\}$.

You can add extra hypotheses which make the conclusion true. For example, if $A\subseteq E$ and $m(A)<\infty$ then $m(E-A)=m(E)-m(A)$. Indeed, since $A\subseteq E$ the (measurable) sets $E-A$ and $A$ are disjoint and have union $E$, so $m(E-A)+m(A)=m(E)$, and since $m(A)<\infty$ we can subtract it to get $m(E-A)=m(E)-m(A)$.

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Yes, mac, thank you so much. m(A) is surely less than infinity since we're considering complex measures, so we don't have to add in that assumption. I actually missed out on saying that A is a subset of E, I missed out on it completely and that is why I got lost in the proof. Thanks again! –  paul May 14 '11 at 17:20
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In general $m(B-A) \neq m(B)-m(A)$. If $B$ and $A$ are disjoint and $m(B)=m(A)>0$, then $m(B)=m(B-A)>0$ yet $m(B)-m(A)=0$.

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