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Suppose A,B are compact metric spaces with fully supported Borel probability measures $m_A$ and $m_B$ correspondingly. Suppose that $A$ is second-countable. Let $f:A\to B$ be a continuous surjection.

  1. Is it true that if $m_A(K) = 1$ for a Borel set $K$ then $m_B(f(K) ) = 1$ ?
  2. If the answer to the first question is no, would it help if I assume in addition that $f$ is Holder continuous (with respect to metrics that are compatible with Borel structures)?

This is a continuation of this question.

The original motivation for the question is the case when $A$ is a Hilbert cube and $B$ is a line segment.

UPD: I have asked a continuation of this question here.

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demitau
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1 Answers1

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This is still not true. Let $A = B = [0,1]$ and let $(q_n)_n$ be an enumeration of $\Bbb{Q} \cap [0,1]$ and $(p_n)_n$ an enumeration of $(\Bbb{Q} + \sqrt{2}) \cap [0,1]$.

Define

$$ m_A (M) = \sum_{n \in \Bbb{N} : q_n \in M} \frac{1}{2^n} $$

and

$$ m_B (M) = \sum_{n \in \Bbb{N} : p_n \in M} \frac{1}{2^n}. $$

Since $\Bbb{Q} \cap [0,1]$ is contained in ${\rm supp} m_A$, we see that $m_A$ has full support. Likewise, $m_B$ has full support.

But if we let $f$ be the identity map (which is certainly Hölder continuous) and $K = \Bbb{Q} \cap [0,1]$, then $m_B (f(K)) = 0$, since $\Bbb{Q}$ and $\Bbb{Q} + \sqrt{2}$ are disjoint.

PhoemueX
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