Prove that $x \mapsto \boldsymbol{\mathrm P} \left[ \mathrm{B}(\boldsymbol\cdot + x ) \in A \right]$ is measurable for a Brownian motion $\mathrm B$ 
Given a (one-dimensional) Brownian motion $\mathrm{B}$ and any $A \in \mathcal B\left(\mathbb R^{[0,1)}\right)$, how may one prove that the map, from $\mathbb R$ to $[0,1]$, given by
$$
x \mapsto \boldsymbol{\mathrm P} \left[ \mathrm{B}(\boldsymbol\cdot + x ) \in A \right] 
$$
is measurable?

The notation
$$
\boldsymbol{\mathrm P} \left[ \mathrm{B}(\boldsymbol\cdot + x) \in A \right]
$$
means that we consider the whole path; or in other words, if the function
$$
t \mapsto \mathrm{B}(t + x) 
$$
is in $A$.
Most grateful for any advice provided!
 A: 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.
A: 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:

*

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


*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.


*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.
