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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.2012-03-23

1 Answers 1

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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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    Thanks to everyone, that was indeed was I was looking for, that terrible measure counting the reals ...2012-03-23