Prove that $x \mapsto \boldsymbol{\mathrm P} \left[ \mathrm{B}(\boldsymbol\cdot + x ) \in A \right]$ is measurable for a Brownian motion $\mathrm B$

A start: If $A$ is of the form $\cap_{k=1}^n\{B(t_k)\in G_k\}$ where $0<t_1<t_2<\cdots <t_n\le 1$ and the $G_k$ are open, then $x\mapsto{\bf P}[B(\cdot+x)\in A]$ is lower semi-continuous (hence Borel measurable) because $B$ is continuous in $t$. Events of this type form a $\pi$-system that generates $\mathcal B(\Bbb R^{[0,1]})$. Now use the Monotone Class Theorem.

As suggested by John Dawkins, we may start with show that the map is measurable when $A$ is of the form $$ \bigcap_{k=1}^n \pi_{t_k}^{-1}(G_k), $$ where $\pi_t$ denotes the projection from $\mathbb R^{[0,1)}$ to $\mathbb R$ given by $$ \pi_t(f) = f(t), $$ and each $G_k$ is an open set in $\mathbb R$.

When $A$ is of the specified form, the map is measurable:

We say that the map $$ x \mapsto \boldsymbol{\mathrm P}\left[\mathrm{B}(\boldsymbol \cdot + x ) \in A\right] $$ is semicontinuous if for any $t$, the set $$ \left\{ x: \boldsymbol{\mathrm P}\left[\mathrm{B}(\boldsymbol \cdot + x ) \in A\right] > t \right\} $$ is open. And we note that a semicontinuous map is Borel-measurable as any Borel-set in $\mathbb R$ may be written as a countable union of sets of the form $(t, \infty)$ and the preimage of a countable union is the countable union of the (individual) preimages.

To prove semicontinuity we need to show that if $$ x_0 \in \left\{ x: \boldsymbol{\mathrm P}\left[\mathrm{B}(\boldsymbol \cdot + x ) \in A\right] > t \right\} $$ then there exists an open interval $(x_1, x_2)$ about $x_0$. Assume that $A$ is of the form $$ A = \bigcap_{k=1}^n\left\{B(t_k)\in G_k\right\}, \quad 0<t_1<\cdots <t_n\le 1, \, G_k \text{ open}. $$ Suppose that $$ \omega \in \left \{ \mathrm{B}(\boldsymbol\cdot + x_0 ) \in A \right \} = \bigcap_{k=1}^n \left \{ \mathrm{B}(t_k + x_0) \in G_k \right \} $$ and that $\omega$ is in the set of probability one such that the paths of $\mathrm{B}$ are continuous. Then, since each $G_k$ is open, by letting $x_1 < x_0 < x_2$ be sufficiently close to $x_0$ we get that for $x \in (x_1, x_2)$ $$ \omega \in \bigcap_{k=1}^n \left \{ \mathrm{B}(t_k + x) \in G_k \right \} = \left \{ \mathrm{B}(\boldsymbol\cdot + x ) \in A \right \}. $$ Hence, for $x \in (x_1, x_2)$ $$ \boldsymbol{\mathrm P}\left[\mathrm{B}(\boldsymbol \cdot + x ) \in A\right]= \boldsymbol{\mathrm P}\left[\mathrm{B}(\boldsymbol \cdot + x_0) \in A\right] > t . $$

Define the class $\mathcal A$ as the collection of set $A \subset \mathbb R^{[0,1)}$ such that, $$ x \mapsto \boldsymbol{\mathrm P}\left[\mathrm{B}(\boldsymbol \cdot + x ) \in A\right] $$ is measurable.

As shown above, the collection of sets $$ \left\{\bigcap_{k=1}^n \pi_{t_k}^{-1}(G_k): \; n \in \mathbb N, t_1, \dots t_n \in [0,1), G_k \subset \mathbb R \text{ open}, \right\} $$ is contained in $\mathcal A$. We also know that this collection of sets forms a $\pi$-system for the Borel-sigma algebra on $\mathbb R^{[0,1)}$. We may use Dynkin's $\pi$-$\lambda$ theorem, by showing that the $\mathcal A$ also is a $\lambda$-system and then conclude that it thus must contain the sigma-algebra generated by the $\pi$-system $\mathcal A$ - i.e. the Borel-sigma algebra.

Let us show that $\mathcal A$ is a $\lambda$-system:

1. clearly $\mathbb R^{[0,1)}$ is in $\mathcal A$ as we may take any $t \in [0,1)$ and consider $$ \mathbb R^{[0,1)} = \pi_t^{-1}(\mathbb R ). $$

2. suppose that $A, B \in \mathcal A$, and $A\subsetneq B$ From the \textit{excision property} of probability measures $$ \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol\cdot + x ) \in B \setminus A \right ] = \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol\cdot + x ) \in B \right ] - \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol\cdot + x ) \in A \right ], $$ hence $$ x \mapsto \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol\cdot + x ) \in B \setminus A \right ] = \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol\cdot + x ) \in B \right ] - \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol\cdot + x ) \in A \right ] $$ is a measurable mapping as the difference of two measurable mappings is a measurable mapping.

3. Suppose that $A_1, A_2, \dots \in \mathcal A$ is a sequence of pairwise disjoint sets. Since measures are countable additive, $$ \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol \cdot + x ) \in \bigcup_{n=1}^\infty A_n \right ] = \sum_{n=1}^\infty \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol \cdot + x ) \in A_n \right ] $$ and hence $$ x \mapsto \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol \cdot + x ) \in \bigcup_{n=1}^\infty A_n \right ] = \sum_{n=1}^\infty \boldsymbol{\mathrm P}\left [ \mathrm{B}(\boldsymbol \cdot + x ) \in A_n \right ] $$ is the limit of a monotone sequence of measurable functions and hence measurable.

We conclude that $\mathcal A$ is a $\lambda$-system and hence contains the Borel sigma algebra.