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If $A$ and $B$ are measurable sets with $\mu(A) < \mu(B)$ , and $P \subset A$ , $Q \subset B$ are measurable sets with $\mu(P) = \mu(Q)$, is it true that $\mu(A-P) < \mu(B-Q)$? It seems this should be true intuitively, but I am not sure about the proof.

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2 Answers 2

up vote 4 down vote accepted

A measure is $\sigma$-additive, and in particular additive: if $X$ and $Y$ are both measurable and disjoint, then $\mu(X\cup Y) = \mu(X)+\mu(Y)$ (in the extended real sense). So you have $$\mu(A) = \mu(P\cup (A-P)) = \mu(P)+\mu(A-P)$$ and $$\mu(B) = \mu(Q\cup(B-Q)) = \mu(Q)+\mu(B-Q)$$

Since $\mu(A)\lt\mu(B)$, then $\mu(A)$ is finite, hence so is $\mu(P)$, hence so is $\mu(Q)$; so from $\mu(P)+\mu(A-P)\lt \mu(Q)+\mu(B-Q)$ and $\mu(P)=\mu(Q)$ you can deduce that $\mu(A-P) \lt \mu(B-Q)$, as desired (simply subtract $\mu(P)$ from both sides of the inequality, since it is a finite quantity).

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+1.5 (.5 for incredibly fast typing) –  t.b. Apr 26 '11 at 21:28
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Hint: $\mu(A - P) + \mu(P) = \mu(A) \lt \mu(B) = \mu(B - Q) + \mu(Q)$.

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