Limit of expectations on increasing sigma algebra converge to expectation on limit sigma algebra

Let $\mathcal{F}_t=\sigma\{B_s:s\le t\}$ be the Brownian filtration and $t_n\nearrow T>0$ and let $Y\in\mathcal{F}_T$. I am having trouble understanding why the martingale convergence theorem implies that $Y_n:=E(Y|\mathcal{F}_{t_n})$ converges to $E(Y|\mathcal{F}_T)$. I understand that there is a random variable $Y_{\infty}$ such that $Y_n\to Y_{\infty}$ but why does this limit has the form $Y_{\infty}=E(Y|\mathcal{F}_T)$?

Also, I am not sure why $Y_n$ is a martingale. I have $E(Y_{n+1}|\sigma(Y_1,\ldots,Y_n))=E(E(Y|\mathcal{F}_{t_{n+1}})|\sigma(Y_1,\ldots,Y_n))$ so it seems that I need to show $\sigma(Y_1,\ldots,Y_n)=\mathcal{F}_{t_n}$ to conclude, but how is this done?

First, $\{Y_n\}$ here is a martingale w.r.t. $\{\mathcal{F}_n\equiv \mathcal{F}_{t_n}\}$ because $\mathcal{F}_n\subset \mathcal{F}_{n+1}$ and $$ \mathsf{E}[Y_{n+1}\mid\mathcal{F}_n]=\mathsf{E}[\mathsf{E}[Y\mid \mathcal{F}_{n+1}]\mid\mathcal{F}_n]=\mathsf{E}[Y\mid \mathcal{F}_n]=Y_n. $$ Second, if $Y_n$ converges in $L^1$ to some random variable $Y_{\infty}$, the latter equals $\mathsf{E}[Y\mid \mathcal{F}_T]$ a.s. because for each $n\ge 1$, $Y_n=\mathsf{E}[Y_{\infty}\mid \mathcal{F}_n]$. (See Lemma 4.6.6 and Theorem 4.6.8 on page 247 here for details.)