Convergence of a sequence of sampled submartingales

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I'd need a review, whether my proof is correct or if I forgot something.

Given:

  • $M$ is a right-continous submartingale (i. e. $\mathbb{E}[M_t | \mathcal{F}_s] \geq M_s$ for $s \leq t$)
  • $(\tau _n )_{n\in \mathbb{N}}$ is a sequence of stoppping times with $\tau_n \searrow \tau$ as well as $\tau, \tau_n \leq N \in \mathbb{N}$ and $\tau _{i+1} \leq \tau _i$ for $i \in \mathbb{N}$ a. s.
  • $A \in \mathcal{F}_{\tau}$

Claim:

  • $\lim _{n \to \infty} \mathbb{E} [M_{\tau_n}\cdot 1_{A}] = \mathbb{E} [M_{\tau}\cdot 1_{A}]$

My approach to prove this:

  1. We have due to the submartingale property: $\mathbb{E} [M_{\tau_n}\cdot 1_A] \leq \mathbb{E} [M_N\cdot 1_A]$
  2. We also have, that $\mathbb{E}[|M_{\tau_n}|]\leq \mathbb{E}[|M_N|]$ and $\mathbb{E}[|M_{\tau}|]\leq \mathbb{E}[|M_N|]$, due to the submartingale property and since these Stopping times are a. s. bounded
  3. We have the integability of $\sup _{n \in \mathbb{N}} |M_{\tau_n}|$ due to 2, which dominates the $M_{\tau _n}$ for each $n$
  4. Using dominated convergence, we get the conclusion.

I'm not sure about points 3 and 4, they seem incorrect to me or at least as if they need a fixing.