Prove or disprove: $t \mapsto \mathbb{P}( X_t \in A)$ is measurable Let $(X_t)_{t \geq 0}$ be an $\mathbb{R}$-valued stochastic process and let the law of $X_t$ be given by $\mathbb{P}_t$, $t \geq 0$. Let $A \in \mathcal{B}(\mathbb{R})$ be a Borel set. I am interested in the measurability of the following mapping
$$
[0, \infty ) \ni t \mapsto \mathbb{P} (X_t \in A) = \mathbb{E}[1_A(X_t)] \in [0, 1]. \tag{1}
$$
under the conditions stated below.

Here is an example showing that this mapping is not measurable in general. Let $N \subset [0, \infty)$ be a set which is not Borel measurable, and let the random variables $X_t$, $t \geq 0$, be such that $\mathbb{P}(X_t =1)=1_N(t)$. Then clearly $t \mapsto \mathbb{P}(X_t = 1)$ is not measurable.

Now assume that:

If $(t_n) \subset [0, \infty)$ is such that $t_n \rightarrow t_0 \in [0, \infty)$, then $\mathbb{P}_{t_n}$ converges weakly to $\mathbb{P}_{t_0}$. This means that for every continuous and bounded function $f : \mathbb{R} \rightarrow \mathbb{R}$ we have
$$
\int_{\mathbb{R}} f(x) \mathbb{P}_{t_n} (dx) = \mathbb{E} [f (X_{t_n})] \overset{n \rightarrow \infty}{\rightarrow} \int_{\mathbb{R}} f(x) \mathbb{P}_{t_0} (dx) = \mathbb{E} [f (X_{t_0})]
$$
The assertion is also true for bounded Lipschitz functions. Does this ensure the measurability?

Is it maybe possible to approximate the indicator function $x \mapsto 1_A(x)$ by suitable continuous functions which will imply the measurability? Alternatively, is it possible to obtain a right-continuous modification of $(X_t)_{t \geq 0}$ under these conditions? If so, this will ensure the joint measurability of $(t, \omega ) \mapsto X_t (\omega)$ and Tonelli's theorem will yield the measurability of the mapping in $(1)$.

If the answer is affirmative, it would be interesting to see if it also applies to the case where $X_t$, $t \geq 0$, are $S$-valued for a sufficiently nice metric space $S$.
 A: The answer is affirmative. Let $C_b(\mathbb{R})$ denote the set of all continuous and bounded functions on $\mathbb{R}$. Then the following claim holds:

Claim. Let $(\mathbf{P}_t)_{t\geq 0}$ be a collection of Borel probability measures such that
$$ t \mapsto \int_{\mathbb{R}} \varphi(x) \, \mathbf{P}_t(\mathrm{d}x) $$
is measurable for each $\varphi \in C_b(\mathbb{R})$. Then the map
$$t \mapsto \mathbf{P}_t(A)$$
is measurable for every $A \in \mathcal{B}(\mathbb{R})$.

Indeed, for each open $U \subseteq \mathbb{R}$, there is a sequence $(\varphi_n) \subseteq C_b(\mathbb{R})$ such that $0 \leq \varphi_n \uparrow \mathbf{1}_U$ at every point of $\mathbb{R}$. So by the monotone convergence theorem,
$$ \mathbf{P}_t(U) = \int_{\mathbb{R}} \mathbf{1}_U(x) \, \mathbf{P}_t(\mathrm{d}x) = \lim_{n\to\infty} \int_{\mathbb{R}} \varphi_n(x) \, \mathbf{P}_t(\mathrm{d}x). $$
Since the map $t \mapsto \mathbf{P}_t(U)$ is the pointwise limit of measurable functions $ t \mapsto \int_{\mathbb{R}} \varphi_n(x) \, \mathbf{P}_t(\mathrm{d}x) $ as $n \to \infty$, it follows that the claim is true for open sets.
Moreover, it is clear that the collection of all Borel sets $A$ for which $t \mapsto \mathbf{P}_t(A)$ is measurable contains $\Omega$ and is closed under complements and disjoint countable unions (hence forms a Dynkin system). So by the $\pi$-$\lambda$ theorem, this collection contains all Borel sets of $\mathbb{R}$. Therefore the claim follows.
