Given a metric space $X$ with a Borel sigma-algebra, the stochastic kernel $K(x,B)$ is such that $x\mapsto K(x,B)$ is a measurable function and a $B\mapsto K(x,B)$ is a probability measure on $X$ for each $x$

Let $f:X\to \mathbb R$. We say that $f\in \mathcal C(B)$ if $f$ is continuous and bounded on $B$.

Weak Feller continuity of $K$ means that if $f\in\mathcal C(X)$ then $F\in\mathcal C(X)$ where $$ F(x):=\int\limits_X f(y)K(x,dy). $$

I wonder if it implies that if $g\in \mathcal C(B)$ then $$ G(x):=\int\limits_Bg(y)K(x,dy) $$ also belongs to $\mathcal C(B)$?

  • $\begingroup$ So in one case you're integrating over all of $X$ and you've got a bounded function that's continuous on $X$. In the other case you're integrating over a Borel subset $B$ and you've got a bounded function that's continuous only on $B$. And in the first case the conclusion includes continuity on all of $X$, and in the second only on $B$. Is that a correct understanding of your question? $\endgroup$ – Michael Hardy Aug 16 '11 at 12:59
  • $\begingroup$ @Michael: exactly. $\endgroup$ – Ilya Aug 16 '11 at 13:09
  • $\begingroup$ @Gortaur To clarify my understanding, does the requirement that $B \mapsto K(x,B)$ is a probability measure on $X$, imply that $K(x, X) = 1$ for all $x \in X$ ? $\endgroup$ – Sasha Aug 16 '11 at 13:40
  • $\begingroup$ @Sasha: yes, otherwise it's just a measure. $\endgroup$ – Ilya Aug 16 '11 at 14:09

No. Take $X = \mathbb{R}$ and $K(x,B) = 1_B(x+1)$. (This corresponds to a process that jumps 1 unit to the right at each step.) Note that $\int f(y) K(x,dy) = f(x+1)$, so $K$ is certainly weak Feller.

However, take $f = 1_{(0, \infty)}$ to be a step function and $B = \mathbb{R} \backslash \{0\}$. Then $f \in C(B)$ by your notation. But I'll let you check that $\int_B f(y) K(x, dy) = 1_{(-1, \infty)}(x)$ which is not in $C(B)$.


To make it clearer, you can re-write $ G(x) = \int\limits_X g(y)\mathbf{1}_B(y) K(x,{\rm d} y)$. Now in general $g{\mathbf 1}_B$ is not continuous anymore so as Nate pointed out you should not expect $G$ to be such.

However, if you take a continuous $g $ with $\overline{{\rm support}(g)} \subsetneq B$ then (I let you :) show this) $g{\mathbf 1}_B$ is still continuous and so is $G$ then.


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