I've a question about a proof in my lecture notes. We want to prove the following theorem.
$M=(M_t)_{t\ge 0}$ be a $(P,F)$-martingale, where $P$ is a probability measure and $F=(\mathcal{F}_t)$ a filtration (right continuous). Further we assume that $M$ has right continuous paths and $\sigma,\tau$ are $F$-stopping times with $\sigma\le \tau$. If $\tau$ is bounded or if $M$ is uniformly integrable, then $E[M_\tau|\mathcal{F}_\sigma]=M_{\sigma}$
The proof consists of 3 steps.
1) Here we conclude: We can assume that $M_t=E[M_N|\mathcal{F}_t]$ for all $t\le N$ with some fixed $N\in [0,\infty]$.
2) Since $\mathcal{F}_\sigma\subset \mathcal{F}_\tau$ and tower property of conditional expectation it's enough to prove: $E[M_N|\mathcal{F}_\tau]=M_\tau$.
3) Here we assume that $\tau$ has only countably many increasingly ordered values $t_n,n\in \mathbb{N}$ and use stopping theorem for discrete time to show $E[M_\tau]=E[M_N]$. For a general $\tau$ we approximate it from "above" and show similarly $E[M_\tau]=E[M_N]$
Now my questions:
to 2): Is this what is meant in 2): if $E[M_N|\mathcal{F}_\tau]=M_\tau$ is true, then
$E[E[M_N|\mathcal{F}_\tau|\mathcal{F}_\sigma]=E[M_\tau|\mathcal{F}_\sigma]$
The LHS is equal $E[M_N|\mathcal{F_\sigma}]$. Can I apply here 1) to get $E[M_N|\mathcal{F_\sigma}]=M_\sigma$? I'm not quiet sure, since in 1) we have a $t$, and here $\sigma$ is a stopping time.
to 3): Why does the Theorem follows if I just show $E[M_\tau]=E[M_N]$?
Thanks in advance for your help.
hulik