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Given two Borel measures $\mu_1$ and $\mu_2$ on $\mathbb R$, is there always a Borel measure $\mu$ on $\mathbb R$ such that

$$ d\mu_1=w_1 d\mu,\qquad d\mu_2=w_2 d\mu, $$ for some functions $w_1$ and $w_2$ on $\mathbb R$ ?

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I've updated the answer according to the comment by @GEdgar - you may be interested. – Ilya Mar 23 '12 at 9:21
up vote 5 down vote accepted

Yes, if both measures are $\sigma$-finite. Then for such measure to exist it should be s.t. $\mu_i$ are absolute continuous w.r.t. $\mu. $ Just take $$ \mu = \mu_1+\mu_2. $$

If at least one of the measures is not $\sigma$-finite then as @GEdgar mentioned, there is a counterexample. Take $\mu_1$ to be Lebesgue measure and $\mu_2$ to be counting measure. Assume that measure $\mu$ exists, then

  1. by considering $1=\mu_2(\{a\})$ prove that $\mu(\{a\})>0$ for any singleton set $\{a\}$;

  2. by considering $1=\mu_1([0,1])$ prove that $w_1(x)>0$ in only countably many $x\in[0,1]$ (say, on the set $K$);

  3. prove that $\int\limits_{[0,1]\setminus K} w_1(x)\mu(dx)= 0$ while $\mu_1([0,1]\setminus K) = 1$.

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Well, he didn't say $\mu_1$ and $\mu_2$ are sigma-finite! What if $\mu_1$ is Lebesgue measure and $\mu_2$ is counting measure (not sigma-finite). Then $\mu = \mu_1+\mu_2$ I understand. But what are $w_1$ and $w_2$? – GEdgar Mar 22 '12 at 14:24
@GEdgar: you're right. I was presuming that Borel measures are defined to be $\sigma$-finite, which is not the case. Does your example then a counterexample for the OP (I understand, that $\mu_1+\mu_2$ doesn't work in this case)? – Ilya Mar 22 '12 at 15:19
Let Student answer that... – GEdgar Mar 22 '12 at 18:55
@GEdgar: thanks, I gave him some hints. – Ilya Mar 23 '12 at 9:21
Thanks to everyone, that was indeed was I was looking for, that terrible measure counting the reals ... – Student Mar 23 '12 at 15:17

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