# Is positive part of the kernel measurable?

Let $(E,\mathscr E)$ be a measurable space and $Q:E\times \mathscr E\to\Bbb [-1,1]$ be a signed bounded kernel, i.e. $Q_x(\cdot)$ is a finite measure on $(E,\mathscr E)$ for any $x\in E$ and $x\mapsto Q_x(A)$ is a measurable function for any set $A\in \mathscr E$.

For any fixed $x$, let the measure $Q^+_x$ be a positive part of the signed measure $Q_x$ as in Hahn-Jordan decomposition. Is it true that $Q^+$ is a kernel, i.e. is the function $x\mapsto Q_x^+(A)$ measurable for any $A\in \mathscr E$? It clearly holds if $Q$ is an integral kernel, i.e. $$Q(x,\mathrm dy) = q(x,y)\mu(\mathrm dy)$$ where $\mu$ is a finite measure on $(E,\mathscr E)$ and $q:E\times E\to \mathbb R$ is a jointly measurable function, but I am interested in the general case.

Any hints oh how to prove that the measurability holds in general, or on how to derive a counter-example.

This question I asked on MSE, but in three weeks no ideas were mentioned there. Neither helped the bounty which has recently finished. Due to this reason, I hope, it's ok to post the question here.

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Sure it's OK to repost here, because you didn't get an answer at MSE, and made a link to the other post. You might also want to link this post at MSE. –  Wolfgang Loehr Aug 22 '12 at 13:14
@Wolfgang: thanks for the suggestion, done. –  Ilya Aug 22 '12 at 13:37
Your questions might be related to theorem 2.8 in "Measurable sets of measures" of Lester Dubins and David Freedman, Pacific J. Math. 14 (1964), 1211-1222. Another helpful reference could be: Lange, Kenneth Decompositions of substochastic transition functions. Proc. Amer. Math. Soc. 37 (1973), 575–580. Anyway, I strongly believe that the answer to your question is affirmative. –  Jochen Wengenroth Aug 22 '12 at 15:15
@Jochen: thank you very much, I'll take a look on these papers - fortunately they are available. –  Ilya Aug 22 '12 at 15:33

I will assume that $E$ is standard Borel.
A kernel is just a measurable map into the space of measures (with a measurable structure generated by evaluations on sets), so it is sufficient to show that the map $\mu \mapsto \mu^+$ is measurable. Actually, there exist fairly general ways to show that "many things are measurable", but here we can do everything "by hands": $\mu^+(A) = \sup_{B \subset A} \mu(B)$, and $B$'s may be restricted to a countable subalgebra, since $\mathcal{E}$ is countably generated.