# Positive part of the kernel

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, but I am interested in the general case.

Update: after three weeks and 1 bounty I decided to post this question also on MO

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It seems like there are partial results here. –  Michael Greinecker Aug 1 '12 at 20:44
@MichaelGreinecker: +1 thanks, I've taken a look on the paper you had referred to. It's interesting - but it seems like the notion of the weak differentiability that is considered there is somewhat stronger than measurability that I need - and due to this reason there are some topological arguments involved. I am looking rather for measure-theoretical approach: it holds that $Q^+$ is a kernel provided it has a density (regardless of the underlying space) - but maybe in general it is not true. –  Ilya Aug 3 '12 at 8:01
By measurable, do you mean Borel measurable? –  Davide Giraudo Aug 15 '12 at 10:15
@DavideGiraudo: yes, I do mean Borel measurability. –  Ilya Aug 15 '12 at 11:04

The following answer is based on Measurable sets of measures by Dubins and Freedman, which was suggested in a comment at the MO version of this question by Jochen Wegenroth.

The result holds true under the assumption that $(E,\mathscr E)$ is countably generated. The algebra generated by a countable set is countable, so we can assume without loss of generality that there is a countable algebra $\mathscr{A}$ with $\mathscr{E}=\sigma(\mathscr{A})$. Since the difference of two measurable functions is measurable, it suffices to show that $Q^+$ is a kernel.

For all $x\in E$ and $B\in\mathscr{E}$, we have $$Q_x^+(B)=\sup_{A\in\mathscr{E}, A\subseteq B} Q_x(A).$$ Let $\lambda\in [0,1]$. We have $$\{x:Q_x(B)^+<\lambda\}=\bigcup _{A\in\mathscr{E}, A\subseteq B}\{x:Q_x(A)<\lambda\}.$$ Now the union on the righthand side is generally over an uncountable set. But for each $x\in E$, $B\in\mathscr{E}$, and $\epsilon>0$ there is $A\in\mathscr{A}$ such that $Q_x(B\triangle A)<\epsilon$. It follows that $$\bigcup _{A\in\mathscr{E}, A\subseteq B}\{x:Q_x(A)<\lambda\}=\bigcup _{A\in\mathscr{A}}\{x:Q_x(A\cap B)<\lambda\}.$$

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Here is another proof, taken from "Markov Chains" by D. Revuz (lemma 1.5 page 190), always under the assumption that $(E, \mathscr{E})$ is countably generated.

The following answer is almost a paraphrase of the proof of Revuz.

By assumption, there is a sequence of finite partitions $\mathscr{P}_n$ of $E$, such that $\mathscr{P}_{n+1}$ is a refinement of $\mathscr{P}_n$, and $\mathscr{E}$ is generated by $\cup_{n \ge 0} \mathscr{P}_n$. For every $x \in E$, there exists a unique $E_n^x \in \mathscr{P}_n$ with $x \in E_n^x$.

Let $x \in E$ be fixed for the moment. Define $\lambda_x$ as the probability measure which is a multiple of $|Q_x|$. (if $Q_x = 0$, choose it as you want)

Then define a function $f_n$ on $E$ by $\displaystyle f_n(y) = \frac{Q_x(E_n^y)}{\lambda_x(E_n^y)}$ if $\lambda_x(E_n^y) > 0$ and $f_n(y) = 0$ otherwise.

By martingale convergence theorem, we have that $f_n$ converges $\lambda_x$-a.s. to the density of $Q_x$ with respect to $\lambda_x$. Hence $f_n^+$ converges $\lambda_x$-a.s. to the density of $Q_x^+$, and since $f_n$ are uniformly bounded, we have for all $A \in \mathscr{E}$ : $$Q_x^+(A) = \lim_n \int_A f_n^+ \, d \lambda_x$$

Now, if $A \in \mathscr{P}_k$, then for all $n > k$, $\int_A f_n^+ \, d \lambda_x = Q_{x,n}^+(A)$, where $Q_{x,n}^+$ is the positive part of the restriction of $Q_x$ to the $\sigma$-algebra generated by $\mathscr{P}_n$. It's easy to see that the map $x \to Q_{x,n}^+(A)$ is measurable, and so is the map $x \to Q_x^+(A)$.

We have hence proven that $x \to Q_x^+(A)$ is measurable for every $A \in \cup_{n \ge 0} \mathscr{P}_n$, and then a Dynkin class argument finishes the proof.

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