Measurability of diagonal set originated from Reflection Principle of Brownian motion Let $B = (B_t)_{t\geq 0}$ be a Brownian motion from probability space $(\Omega, \mathcal{F}, P)$ to $(C(\mathbb{R}^+, \mathbb{R}), \mathcal{C}$), ($\mathcal{C}$ is the  usual Borel $\sigma$-algebra generated by the topology of compact convergence) and stopping time $T_a = \inf \{t \geq 0, B_t = a \}$. Let $t>0,b \in \mathbb{R}$, we have $$ P(T_a\leq t, W_{t-T_a} \leq b) = P(T_a \leq t, -W_{t-T_a} \leq b), \quad (*)$$ where $W_t = B_{t+T_a}-B_{T_a}$ is a Brownian Motion independent of $\mathcal{F}_{T_a}$. This can be proved in many ways, e.g, using Fubini Theorem. However, I have encountered another view from the book "Stochastic Calculus" by Baldi, page 69, stating that we can prove $(*)$ by rewriting it as
$$ P((T_a, W) \in A) =  P((T_a, -W) \in A),$$ where $A = \{ (s,f) \in \mathbb{R} \times C(\mathbb R^+, \mathbb{R}) \}: s\leq t, f(t-s) \leq b \}$, and by using the independence of $T_a$ and $W$. This point is clear but what really interests me is how to prove $A$ is measurable in the product $\sigma$-algebra, i.e, $A \in \mathcal{B}(\mathbb{R}^+) \otimes \mathcal C$. It is also clear that we only need to prove $A \in \mathcal{B}([0,t]) \otimes \mathcal C$. I have two approaches for this (which I will write in a separate answer under this post): the first involves seeing $A$ as a diagonal set and the second involves proving $A$ as a preimage of a continuous function $g: [0,t]\times C(\mathbb R^+, \mathbb R) \mapsto \mathbb R$.
What I hope to gain from this post is to know whether there are other better, shorter or easier-to-understand approaches in your opinions? Also, are my proofs wrong? I personally think mine are too long, especially the first one and it may not be very useful.
 A: First approach: A is the countable intersection of the unions of diagonal blocks
Let $\mathcal A_c :=  \{ f: f(t-c) \leq b\}$. Observe that we can write
$$ A = \bigcup_{c \in [0,t]} \{ c \}\times \mathcal A_c. $$
For each $n \geq 1$ and $i =0, 1,...,2^n-1$, define
$$ X^i_n = \Big[\frac{it}{2^n},\frac{(i+1)t}{2^n} \Big] \times \bigcup_{c \in \big[\frac{it}{2^n},\frac{(i+1)t}{2^n} \big]} \mathcal A_c, $$
and let $U_n: = \cup_{j=1}^n X^j_n$, the union of the diagonal blocks (like rectangles sharing 2 vertices on the diagonal). We can prove that $\bigcup_{c \in [p,q]} \mathcal A_c$  is measurable for all $p\leq q\leq t$
. Also $A \subset U_n$ since $(s,f) \in A $ implies $(s,f) \in \{s \} \times \mathcal A_s$. So $A \subset \cap_{n\geq 1} U_n$.
Now I would like to do the reverse inclusion, i.e, $\cap_{n\geq 1}U_n \subset A $. Suppose that $(s,f) \in \cap_{n\geq 1}U_n$, which implies $\forall n \geq 1, \exists j=j(n,s,f) $ such that $(s,f) \in X^{j(n)}_n$. Since if $s \in \Big[\frac{j(n)t}{2^n},\frac{(j(n)+1)t}{2^n} \Big]$ then $s$ can belong to one of (or both) $\Big[ \frac{2j(n)t}{2^{n+1}},\frac{(2j(n)+1)t}{2^{n+1}} \Big] $  and $\Big[ \frac{(2j(n)+1)}{2^{n+1}},\frac{(2j(n)+2)t}{2^{n+1}} \Big] $, we choose $j(n+1)$ to be $2j(n)$ or  $2j(n)+1$. With this construction $(X^{j(n)}_{n})_{n\geq 1}$ is a decreasing sequence of sets and $(s,f) \in \cap_{n\geq 1}X^{j(n)}_n $.
Noticing that $(X^{j(n)}_n)_{n \geq 1}$ is a decreasing sequence of sets and their intersection is
$$ \bigcap_{n\geq 1}X^{j(n)}_n =\bigcap_{n\geq 1} \Big[\frac{j(n) t}{2^n},\frac{(j(n)+1)t}{2^n}\Big] \times \bigcap_{n\geq 1} \bigcup_{c \in \Big[\frac{j(n) t}{2^n},\frac{(j(n)+1)t}{2^n}\Big]}\mathcal A_c $$
so $\{s\} = \bigcap_{n\geq 1} \Big[\frac{j(n) t}{2^n},\frac{(j(n)+1)t}{2^n}\Big]$ as they are a size-decreasing-to-0 sequence of compact intervals. As a consequence, $\mathcal A_s  \subset \bigcup_{c \in \Big[\frac{j(n) t}{2^n},\frac{(j(n)+1)t}{2^n}\Big]}\mathcal A_c $ for all $n\geq 1$ and hence $\mathcal A_s \subset \bigcap_{n\geq 1}\bigcup_{c \in \Big[\frac{j(n) t}{2^n},\frac{(j(n)+1)t}{2^n}\Big]}\mathcal A_c $. Now, I claim that indeed $\mathcal A_s = \bigcap_{n\geq 1}\bigcup_{c \in \Big[\frac{j(n) t}{2^n},\frac{(j(n)+1)t}{2^n}\Big]}\mathcal A_c $. Suppose $f \in \bigcap_{n\geq 1}\bigcup_{c \in \Big[\frac{j(n) t}{2^n},\frac{(j(n)+1)t}{2^n}\Big]}\mathcal A_c $, which means $\forall n \geq 1, \exists c_n \in \Big[\frac{j(n) t}{2^n},\frac{(j(n)+1)t}{2^n}\Big]$, such that $f(t-c_n) \leq b$, but this means $f(t-s) = \lim f(t-c_n) \leq b $. So $f \in \mathcal A_s$ and the equality follows. As a consequence, $(s,f) \in \{s\} \times \mathcal A_s \subset A$. Hence, $A = \cap_{n\geq 1} U_n$, measurable.
Edit: Found a way to fix the problem, I changed the texts accordingly.
Second Approach: A is a preimage of a continuous function
Let $g: [0,t]\times C(\mathbb R^+, \mathbb R) \mapsto \mathbb R$ be defined as $g(s,f) = f(t-s)$. We equipped the space $[0,t]\times C(\mathbb R^+, \mathbb R)$ with the metric $d_\infty = \max \{|s-s'|, \|f-f'\|_{\infty}\} $. With this metric, the space $[0,t]\times C(\mathbb R^+, \mathbb R)$ is separable and the Borel $\sigma$-algebra generated by the open sets is equivalent to Borel $\sigma$-algebra generated by the open balls. Moreover, they are the same as the product $\sigma$-algebra $\mathcal B[0,t] \otimes \mathcal C$. This can be seen from
$$d_\infty((s,f),(s',f')) < \epsilon \iff |s-s|< \epsilon, \| f-f' \|_\infty < \epsilon $$
It remains to show that $g$ is continuous. Indeed,  as $(s,f)$ gets close to $(s',f')$
$$ |g(s,f) -g(s',f)| = |f(t-s) -f'(t-s') | \leq | f(t-s) -f(t-s')| + |f(t-s') -f'(t-s')| \rightarrow 0. $$
From this I can see that $g^{-1}(-\infty,b) = A$ and hence $A \in \mathcal B[0,t] \otimes \mathcal C \subset \mathcal B(\mathbb R^+) \otimes \mathcal C$. It seems to me that this approach is more natural than the first one. I look forward to have your opinions.
