# A question about stochastic kernels and invariant measures

Suppose that $E$ is a metric space, let $\mathcal{B}_E$ denote the set of its Borel subsets and suppose that $\mu$ is a probability measure on $(E,\mathcal{B}_E)$. In addition, suppose that $p:E\times \mathcal{B}_E\to[0,1]$ is a stochastic kernel such that

• For any $B\in\mathcal{B}_E$, $p(\cdot,B):E\to[0,1]$ is a measurable map.
• For any $x\in E$, $p(x,\cdot):\mathcal{B}_E\to[0,1]$ is a probability measure on $(E,\mathcal{B}_E)$.

It can be shown (for example, Lemma 2 on page 11 of this) that for any bounded measurable function $f$

$$\int \left(\int f(y)p(x,dy)\right)\mu(dx)=\int f(x)\nu(dx),\quad\quad(*)$$

where $\nu$ is the probability measure on $(E,\mathcal{B}_E)$ defined by

$$\nu(B):=\int p(x,B)\mu(dx).$$

My question is: If $1\leq q <\infty$ Is it ever true that $(*)$ holds for any $f\in L^q(\mu)$ if $\mu$ is an invariant probability measure of $p$? That is, if for any $B\in \mathcal{B}_E$

$$\mu(B)=\int p(x,B)\mu(dx).$$

The reason I think (hope!) there might be something to the above is that, under some mild conditions on $E$ and $\mu$, one can show that the bounded linear operator on the space of bounded measurable functions defined by $p$ can be extended to $L^q(\mu)$.

By linearity, we can assume that $f$ is non-negative. Indeed, in order to do that, we have to be sure that $\int f(y)p(x,dy)$ is finite for almost every $x$ when $f$ is a non-negative element of $\mathbb L^1(\mu)$. Define $g_n(x):=\int f(y)\chi_{\{y, f(y)\geqslant n\}}p(x,dy)$. Using invariance of $\mu$, we have $$\int f(y)\chi_{\{y, f(y)\geqslant n\}}\mathrm d\mu=\iint f(y)\chi_{\{y, f(y)\geqslant n\}}p(x,dy)\mu(dx),$$ hence $\lVert g_n\Vert_{\mathbb L^1(\mu)}\to 0$. Consequently, we can extract a subsequence $(g_{n_k})_k$ converging $\mu$-almost everywhere to $0$. In particular, for $\mu$-almost every $x$, we have finiteness of $\int f(y)p(x,dy)$.

There is a sequence of non-negative bounded measurable functions which converges pointwise to $f$ (more than almost everywhere), say $(f_n)_{n\geqslant 1}$. We thus have for each $n$, $$\int\left(\int f_n(y)p(x,dy)\right)\mu(dx)=\int f_n(x)\mu(dx).$$ Using monotone convergence theorem, the RHS converges to $\int f(x)\mu(dx)$ as $n$ goes to infinity.

For the LHS, define $g_n(x):=\int f_n(y)p(x,dy)$: this sequence is pointwise non-decreasing to $\int f(y)p(x,dy)$ hence we conclude using two times the monotone convergence theorem.

• Hi, thanks for the post. Quick question, to go from non-negative functions to $f\in L^q(\mu)$ do we not require that $$\int \max\{f,0\}p(x,dy)+\int \min\{f,0\}p(x,dy)$$ is well defined (i.e. ensure that we don't end up with $\infty-\infty$)? How can we guarantee this, simply knowing that $f\in L^q(\mu)$? Or I'm I missing something?
– jkn
Commented Jan 11, 2014 at 14:10
• I've added details. Commented Jan 11, 2014 at 15:28
• Sorry, one last subtlety (if you have the time). From the above we have that if $f\in L^p(\mu)$, then $g(x):=\int f(y)p(x,dy)$ is defined $\mu$-almost everywhere (vs if $f$ was bounded and measurable it's straightforward to show that $g$ is defined everywhere and measurable). So if $f\in L^q(\mu)$, does it make sense to be integrating $g$ with respect to $\mu$?
– jkn
Commented Jan 11, 2014 at 17:45
• Actually, we have finiteness of the LHS. Pick indeed $n$ such that the RHS of the first displayed equation of the answer is small than $1$. Commented Jan 11, 2014 at 17:53