Absolute continuity of the kernel

Let $(X,d)$ be a metric space and $K:X\times \mathcal B(X)\to[0,1]$ is a stochastic kernel on $X$. We call this kernel absolute continuous with respect to a measure $\mu:\mathcal B(X)\to\mathbb R_{\geq0}$ if for any $B\in\mathcal B(X)$ such that $\mu(B) = 0$ holds $K(x,B) = 0$ for all $x\in X$.

What a necessary and sufficient conditions for such measure $\mu$ to exist given a kernel $K$?

• I think that if $K(x,\cdot)$ is a measure for each $x$, then $\mu(\cdot)=\sum_{x\in X}K(x,\cdot)$ is a measure, and each $K(x,\cdot)$ is absolutely continuous with respect to $\mu$. – Jason Swanson Jun 30 '11 at 9:45

Suppose that $K(x,\cdot)$ is a measure for each $x$. Define $$\mu(B) = \sum_{x\in X}K(x,B).$$ This possibly uncountable sum can be understood either as the supremum over all finite subsums, or as the integral with respect to counting measure on $X$. Those are equivalent definitions. The latter is useful, so that we may apply tools of integration theory, such as Fubini/Tonelli.

To check that $\mu$ is a measure, first note that $\mu(\emptyset)=0$. Let $\{B_n\}$ be pairwise disjoint. Then \begin{align*} \mu\bigg(\bigcup_{n=1}^\infty B_n\bigg) &= \sum_{x\in X}K\bigg(x,\bigcup_{n=1}^\infty B_n\bigg)\\ &= \sum_{x\in K}\sum_{n=1}^\infty K(x,B_n)\\ &= \sum_{n=1}^\infty\sum_{x\in X} K(x,B_n)\\ &= \sum_{n=1}^\infty\mu(B_n). \end{align*} Reversing the order of summation is justified by Tonelli's theorem.

Finally, if $\mu(B)=0$, then $K(x,B)=0$ for all $x$. Hence, $K$ is absolutely continuous with respect to $\mu$.

Edit:

To address the comments, let us suppose that $x\mapsto K(x,U)$ is Borel measurable for each open $U$. We will show that this implies $x\mapsto K(x,B)$ is Borel measurable for every $B\in\mathcal{B}(X)$.

Let $f_B(x)=K(x,B)$. Let $$\mathcal{D} = \{B\in\mathcal{B}(X): f_B\text{ is Borel measurable}\}.$$ Since $X$ is open, we have $X\in\mathcal{D}$. If $A,B\in\mathcal{D}$ and $B\subset A$, then $f_{A\setminus B}=f_A-f_B$ is measurable, which implies $A\setminus B\in\mathcal{D}$. And if $\{A_n\}_{n=1}^\infty$ is an increasing sequence in $\mathcal{D}$ with $A=\bigcup_{n=1}^\infty A_n$, then $f_A=\sup_n f_{A_n}$ is measurable, which shows that $A\in\mathcal{D}$. In all, we have shown that $\mathcal{D}$ is a Dynkin system that contains the open sets. So by the $\pi$-$\lambda$ theorem, $\mathcal{D}=\mathcal{B}(X)$.

If $\nu$ is a measure on $(X,\mathcal{B}(X))$, then we may now define $$\mu(B) = \int_X K(x,B)\nu(dx).$$ The proof that $\mu$ is a measure is the same as what was done above for counting measure. However, now if $\mu(B)=0$, then we may only conclude that $K(x,B)=0$ for $\nu$-a.e. $x$. A word of warning: the exceptional null set may depend on $B$. More specifically, we can say that $$\forall B,\;\mu(B)=0 \quad\Rightarrow\quad(K(x,B)=0\text{ for \nu-a.e. x}).$$ But we can *not* say that $$(\forall B,\;\mu(B)=0 \quad\Rightarrow\quad K(x,B)=0)\text{ for \nu-a.e. x}.$$ In fact, there may not be a single $x\in X$ such that $K(x,\cdot)$ is absolutely continuous with respect $\mu$.

For example, let $X=\mathbb{R}$ and $K(x,B)=1_B(x)$. In other words, $K(x,\cdot)=\delta_x$, the point mass centered at $x$. Let $\nu=m$, where $m$ is Lebesgue measure. Then $$\mu(B) = \int_{\mathbb{R}}1_B(x)m(dx) = m(B).$$ That is, $\mu=m$. It is true that for each fixed $B$ with Lebesgue measure zero, $\delta_x(B)=0$ for Lebesgue a.e. $x$. However, there is not a single $x\in\mathbb{R}$ such that $\delta_x(B)=0$ for every Lebesgue null $B$.

• thank you, could we think about conditions on the measure $\nu$ to use it instead of the counting measure, or you prefer to see it as a new question? – Ilya Jun 30 '11 at 10:05
• If $x\mapsto K(x,B)$ is measurable for each $B$, then you could define $\mu$ that way. But in the end, given $\mu(B)=0$, all you could conclude is that $K(x,B)=0$ for $\nu$-a.e. $x$. – Jason Swanson Jun 30 '11 at 10:11
• what if $K(x,B)$ is continuous for all open $B$? – Ilya Jun 30 '11 at 10:25