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](http://en.wikipedia.org/wiki/Hahn_decomposition_theorem). 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.




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This question I asked on [MSE](https://math.stackexchange.com/questions/177631/positive-part-of-the-kernel), 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.



Is positive part of the kernel measurable?
I will assume that $E$ is [standard Borel](http://en.wikipedia.org/wiki/Borel_space#Standard_Borel_spaces_and_Kuratowski_theorems).


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.


