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This question concerns Feller Markov kernels, similar to Vanessa's question.

Terminology

By 'Markov kernel' $N:E\to F$, we adopt exactly the same definition as Vanessa, with the exception that $E,F$ are assumed to only be separable metric spaces. And we define '$N$ being Feller' as in Dellacherie and Meyer (1988) page 6, which reads:

The kernel $N:E\to F$ is said to be Fellerian if the map $E\ni x\mapsto N(x,-)$ is continuous for the narrow topology ...

In other words, $Nf:E\to\mathbb{R}$ is continuous for all $f\in C_b(F)$, where $$ Nf(x):=\int_FN(x,dy)f(y). $$

Question

Dellacherie and Meyer remark on this definition:

When $F$ is locally compact, to say that $N$ is Fellerian reduces to saying that the function $N1$ is continuous, and that the function $Nf$ is continuous on $E$ for continuous functions $f$ on $F$ of compact support (Bourbaki, Intégr., chap. IX $\S$5, prop. 9).

I assumed 'reduces to' meant 'equivalent to', and I've been trying to prove it without success. While the necessity is easy to prove (this direction is even treated as exercise 5.13 in Revuz (1984) p.38, and I'll provide a sketch in the comment section), I doubt that $N1\in C(E)$ and $N(C_c(F))\subset C(E)$ are sufficient conditions for $N$ to be Fellerian.

Discussion

The cited proposition in Bourbaki (2004) concerns uniform approximation of bounded measurable functions via simple functions. The exact statement reads:

(Let $X$ be a locally compact space and $F$ a topological vector space) If $F$ is a metrizable compact space, then every measurable mapping $f$ of $X$ into $F$ is the uniform limit (on all of $X$) of a sequence of measurable step functions.

One might attempt to establish continuity of $Nf$ for an arbitrary $f\in C_b(F)$ via the uniform convergence: $$ \|Nf-Nf_n\|_\infty\le\|f-f_n\|_\infty\to0 $$ ensuring that $Nf_n\in C(E)$. However, this approach faces two obstacles:

  1. If, for instance, $E=\mathbb{R}$ and $f=1_{[0,\infty)}$, we cannot uniformly approximate $f$ using only $1$ and the elements of $C_c(E)$;
  2. Since $Nf$ is not necessarily bounded, $Nf$ might not be the uniform limit of $Nf_n$'s.
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  • $\begingroup$ If $N$ is fellerian, then for all $f\in C_b(F)$, $x\mapsto Nf(x)$ is a continuous function on $E$. This seems obvious when we decompose it into $x\mapsto N(x,-)\mapsto (N(x,-),f)=Nf(x)$, which is a composition of two continuous mappings. To finalize the proof, we take $f=1$ and $f\in C_b(F)\subset C_b(F)$. $\endgroup$ Commented Mar 24, 2024 at 7:17
  • $\begingroup$ I noticed there was error in 2. $Nf$ is bounded if $f$ is bounded. $\endgroup$ Commented Mar 24, 2024 at 8:45

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Let $\langle\mu_n\rangle$ be a sequence of probability measures on $E$ converging weakly to $\mu$, and $f\in C(F)$. We can assume without loss of generality that $f$ takes values between $0$ and $1$.

Suppose now that there exists for each $\epsilon>0$ nonnegative compactly supported functions $g$ and $h$ such that $g\leq f\leq 1-h$ and $\mu N g\geq \mu N(1-h)-\epsilon$. Then, by a sandwich argument, $\lim_n \mu_n N f=\mu Nf$.

It remains to show that such functions exist. Using the local compactness of $F$ and the fact that metric spaces are perfectly normal, one can find a compactly supported function $d$ with values between $0$ and $1$ such that $\mu N(1-d)<\epsilon/2$. One can then take $g=f\wedge d$ and construct $h$ similarly.

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  • $\begingroup$ Thank you for answering! Just to make sure I understand your proof correctly, 1. do you mean $\lim_n\mu_nNf=\mu Nf$? and 2. I guess this statement imply weak convergence $\mu_nN\Rightarrow\mu N$, but how does this related to the original problem: $E\ni x\mapsto N(x,-)\in\mathcal{P}(F)$ is continuous? $\endgroup$ Commented Mar 24, 2024 at 12:57
  • $\begingroup$ 1. Yes, thanks (corrected it). 2. For separable metric spaces, weak convergence is metrizable. So, continuity is equivalent to sequential continuity. And, the set of Dirac measures on $E$ is closed in the topology of weak convergence. So just take the $\mu_n$ to be Dirac measures. $\endgroup$ Commented Mar 24, 2024 at 13:07
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    $\begingroup$ Yes! I finally grasped it. Essentially, you also approximate $f$ via a certain sequence $\{f_m\}\subset C_c(F)$. However, in this case, we don't need uniformity because $(\mu_nN)f$ is an integral. Therefore we can apply the bounded convergence theorem to obtain $(\mu_nN)f_m\to(\mu_nN)f$. This is a part of how you derived $\lim_n\mu_nNf=\mu Nf$, right? $\endgroup$ Commented Mar 25, 2024 at 6:17
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    $\begingroup$ Yes, exactly. The approximation is allowed to depend on the sequence. $\endgroup$ Commented Mar 25, 2024 at 7:28

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