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Is every analytic set $A$ in, say, $I=[0,1]$, the projection of a Borel set $B$ in, say, $[0,1] \times [0,1]$, $A = \pi_1(B)$, with the following property: For every regular Borel probability measure $\mu$ on $[0,1]$ and (EDIT) $\mu(A)=1$, there is $y \in [0,1]$ s.t.

$ \int 1_B(x,y) \mu(dx) > 0 $

where $1_B(\cdot,\cdot)$ is the indicator function of $B$?

Any results that show that analytic sets are projections of sets with sections 'large' in some sense are appreciated as well. Answers that use axioms in addition to ZFC (in particular, $V=L$) are also useful.

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    $\begingroup$ Just some comments on large and small sections. In the opposite direction some sections must be small in some sense, for example if $A\subseteq [0,1]^2$ is Borel and $A_x=\{y\in [0,1]\mid (x,y)\in A\}$ is nonmeager for every $x\in [0,1]$ or if $\mu(A_x)>0$ for all $x\in[0,1]$ and some Borel probability measure $\mu$, then $\pi_1(A)$ must be Borel. In the right direction some section must be big in some sense, for example if $A\subseteq[0,1]^2$ is Borel and $A_x$ is $\sigma$-compact for all $x$, then $\pi_1(A)$ is Borel. $\endgroup$ Aug 2 at 13:47
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    $\begingroup$ Sorry I mean that $\mu(A)=1$, added now, thnx! @Arno $\endgroup$
    – John Levy
    Aug 2 at 18:59
  • $\begingroup$ Could you clarify that you intend the quantifier order that you have stated, namely, $\forall A\exists B\forall\mu\exists y$? Do you already know what happens for the easier requirement $\forall A\forall\mu\exists B\exists y$? $\endgroup$ Aug 2 at 19:35
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    $\begingroup$ @JoelDavidHamkins For the easier quantifier order, the answer is definitely yes. Pick a compact subset $A'$ of $A$ with positive $\mu$-measure, and just add change $B$ to $(A' \times \{0\}) \cup B$. $\endgroup$
    – Arno
    Aug 2 at 20:26
  • $\begingroup$ @AlessandroCodenotti Thnx, yeah, I knew the measure one - wasn't sure about the category one but good to know as well. $\endgroup$
    – John Levy
    Aug 2 at 20:34

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