Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
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

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
That is essentially the proof Dubins and Freedman give in the paper Jochen mentioned. The essamption that the underlying measurable space is standard Borel can be weakened to the assumtion of being countably generated. –  Michael Greinecker Sep 4 '12 at 10:37
@Michael: thanks for the comment –  Ilya Nov 30 '12 at 16:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.