Let $ (X,\mu) $ be a standard measure space - so that we may assume that $X$ is the unit interval $[0,1]$ with the Borel $\sigma$-algebra. Consider $X \times X$ with the product measure $\mu \times \mu $ defined on the product $\sigma$-algebra.

Let $f$ and $g$ be two functions defined on $X \times X$ such that:

  1. $f$ is measurable.
  2. For a.e. $x$, the function $y \to g(x,y)$ is measurable.
  3. For every measurable subset $E\subset X \times X$, we have $$\int \chi _E \cdot f(x,y) d\mu(y) = \int \chi_E \cdot g(x,y) d\mu(y) $$ for a.e. $x$.

Is this data sufficient to imply that $g$ is measurable as well? and that $f=g$ a.e.?

This is a (hopefully more interesting) variant of my previous question Measurable functions on product measures, which was answered in the negative by @Nate Eldredge (but the same example does not work in this case).


As mentioned here, take a bijection of $[0, 1]$ whose graph has outer area one. Let $g$ be the characteristic function of this graph. Let $f$ be the constant function 0. Now all of your 1, 2, 3 above hold but the conclusions fail.

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    $\begingroup$ If you are the same hot queen, then you might want to visit this page. $\endgroup$ – Asaf Karagila May 1 '13 at 18:26
  • $\begingroup$ I could not login from a different computer and posting an answer here requires you to give a name so I used my previous user name. $\endgroup$ – hot_queen May 1 '13 at 18:46
  • $\begingroup$ Thank you hot_queen. I understand how such an automorphism would be a counter-example. But could you please give a hint or a reference to how to construct it? $\endgroup$ – the_lar May 2 '13 at 6:15
  • $\begingroup$ Try using transfinite induction. $\endgroup$ – hot_queen May 2 '13 at 13:15
  • $\begingroup$ The point of my previous comment was that you can have the two accounts merged, either yourself (if you still have access to the other computer) or by sending a request to the SE team. $\endgroup$ – Asaf Karagila May 2 '13 at 14:19

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