A stochastic process with right continuous paths and usual conditions to the adapted filtration is progressively measurable I'm reading Stochastic Processes by Richard F. Bass and got stuck by Exercise 1.3 for a while.

Let $\{\mathcal{F}_t\}$ be a filtration satisfying the usual conditions (complete and right-continuous) and let $\mathcal{B}[0,t]$ be the Borel $\sigma$-field on $[0,t]$. A real-valued stochastic process $X$ is progressively measurable if for each $t\ge 0$, the map $[0,t]\times \Omega\colon (s,\omega)\mapsto X(s,\omega)$ is $(\mathcal{B}[0,t]\otimes\mathcal{F}_t)$-measurable.
Show that if $X$ is adapted to $\{\mathcal{F}_t\}$ and has right-continuous paths, then $X$ is progressively measurable.

My attempt. For each $n\in\mathbb{N}$, define on $[0,\infty)\times \Omega$ a function $$Y^{(n)}(s,w)=\sum_{k=0}^\infty X((k+1)/2^n,\omega)1_{[k/2^n,(k+1)/2^n)}(s).$$
Let $t\ge 0$ be given. It is clear that the map $[0,t]\times \Omega\colon (s,\omega)\mapsto Y^{(n)}(s,\omega)$ is $(\mathcal{B}[0,t]\otimes \mathcal{F}_{t+2^{-n}})$-measurable. Let $\hat{k}(n)$ be the unique integer such that $$\frac{\hat{k}(n)}{2^n}\le s < \frac{\hat{k}(n)+1}{2^n}.$$ Then $$Y^{(n)}(s,\omega)=X((\hat{k}(n)+1)/2^n,\omega)\underset{n\to\infty}{\longrightarrow}X(s,\omega)$$ since $(\hat{k}(n)+1)/2^n\downarrow s$ by construction and $X$ has right-continuous paths. Hence, the map $[0,t]\times \Omega\colon (s,\omega)\mapsto X(s,\omega)$ is $\bigcap_{n\in\mathbb{N}}(\mathcal{B}[0,t]\otimes \mathcal{F}_{t+2^{-n}})$-measurable. The proof would be finished if I can show $$\bigcap_{n\in\mathbb{N}}(\mathcal{B}[0,t]\otimes \mathcal{F}_{t+2^{-n}})\subseteq \mathcal{B}[0,t]\otimes \mathcal{F}_{t}.$$
I am stuck here. I haven't used the usual conditions, especially the right-continuity of $\{\mathcal{F}_t\}$. Is this the right direction?
Thank you!
 A: The following is a longer version of a standard argument (see e.g. Karatzas and Shreve, Proposition 1.13). Define
$$Y^{(n)}(\omega,s)=\begin{cases}X_{t(k+1)2^{-n}}(\omega)& s \in (tk2^{-n},t(k+1)2^{-n}],\,k=\{0,1,...,2^n-1\}\\
X_0(\omega)&s=0\end{cases}$$
We write more explicitly
$$Y^{(n)}(\omega,s)=X_0(\omega)\mathbf{1}_{\{0\}}(s)+\sum_{1\leq k<2^n}X_{t(k+1)2^{-n}}(\omega)\mathbf{1}_{(tk2^{-n},t(k+1)2^{-n}]}(s)$$
(i) To see that this is $\mathscr{F}_t\otimes \mathscr{B}[0,t]$-measurable, first note that all $\omega \mapsto X_{t(k+1)2^{-n}}(\omega)$ are $\mathscr{F}_t$-measurable functions, since $X$ is adapted, and $s \mapsto \mathbf{1}_{(tk2^{-n},t(k+1)2^{-n}]}(s)$ are $\mathscr{B}[0,t]$-measurable functions. Also for any $A,B \in \mathscr{B}(\mathbb{R})$:
$$\begin{aligned}\{(\omega,s):X_{t(k+1)2^{-n}}(\omega)\in A\}&=\{\omega: X_{t(k+1)2^{-n}}(\omega)\in A\}\times [0,t]\in \mathscr{F}_t\otimes \mathscr{B}[0,t]\\
\{(\omega,s):\mathbf{1}_{(tk2^{-n},t(k+1)2^{-n}]}(s)\in B\}&=\Omega \times \{s:\mathbf{1}_{(tk2^{-n},t(k+1)2^{-n}]}(s) \in B\}\in \mathscr{F}_t\otimes \mathscr{B}[0,t]\end{aligned}$$
So the functions $(\omega,s)\mapsto Y^{(n)}(\omega, s)$ are linear combinations of $\mathscr{F}_t\otimes \mathscr{B}[0,t]$-measurable functions, and are thus measurable.
(ii) The sequence $Y^{(n)}(s,\omega)$ approaches $X_s(\omega)$ from the right and the right continuity guarantees that the limit is $X_s(\omega)$. The limit of a sequence of measurable functions is measurable whenever it exists; we then have that $X_s(\omega)$ is also $\mathscr{F}_t\otimes \mathscr{B}[0,t]$-measurable. But this means that $X_s(\omega)$ is $\mathscr{F}_t$-progressively measurable.
