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.