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Suppose A,B are compact metric spaces with Borel probability measures $m_A$ and $m_B$ correspondingly. Let $f:A\to B$ be a continuous surjection. Is it true that if $m_A(K) = 1$ for a Borel set $K$ then $m_B(f(K) ) = 1$ ? The fact that $f(K)$ is Borel is discussed in Continuous images of open sets are Borel?

UPD: Suppose additionally that $A$ is second-countable (for the link I have attached to work).

I asked continuations of this question here and here.

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demitau
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    I don't see how the question you linked implies that $f(K)$ is Borel. – PhoemueX Mar 05 '15 at 22:25
  • Yes, you are right, I have to suppose additionally that A is second countable. The question is still interesting for me under this assumption. – demitau Mar 06 '15 at 07:26

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Take $A = B = \{1,2\}$ with the discrete topologies and $m_A, m_B$ the unique probability measures with $m_A (\{1\})=1$ and $m_B(\{2\})=1$ and $f(1)=1$, $f(2)=2$.

This yields a counterexample by taking $K=\{1\}$.

PhoemueX
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  • Thank you, this solves my original question. Is it possible to construct a counterexample in the case if $m_A$ and $m_B$ are fully supported?

    Or I should better ask it as a separate question?

     – demitau Mar 06 '15 at 07:38
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    @demitau separate question – hunter Mar 06 '15 at 08:05