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Suppose $u$ and $v$ are measures on a measurable space $E$. Further suppose $u$ is finite and absolutely continuous with respect to v ($v(S)=0 \implies u(S)=0$). The problem is: Show that $\forall \epsilon>0 \;\;\;\exists \delta: v(S)<\delta \implies u(S)<\epsilon.$

I've tried the following:

Fix $\epsilon>0$.

Let $S_n=\{measurable\;\; B\in E|u(B)>\epsilon, v(B)<1/n\}$.

If $S_n$ is finite for some n we are done ($v(S)=0 \implies u(S)=0$).

Now what I am trying to show is that the other case, $S_n$ is infinite $\forall n\in\mathbb{N}$, leads to a contradiction. Any suggestions?

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1 Answer 1

up vote 2 down vote accepted

I think that if you change the 1/n into $1/(2^n)$ and choose a sequence $B_n\in S_n$ such that $v(B_n)<2^{-n}$, and $u(B_n)>\epsilon$, and consider $B=\limsup B_n=\bigcap(\bigcup B_n)$, you might get your contradiction.

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