I was reading some Probability Theory notes and I saw a proof of this result :
Lemma : Let ${(\Omega,\mathcal{F},\mathbb{P})}$ be a probability space equipped with the (discrete) filtration $(\mathcal{F}_n)$. For any stopping time $\tau$ on that probability space, we define the stopped $\sigma$-algebra as ${\mathcal{F}_\tau:=\{A\in\mathcal{F}:\forall n,\{\tau=n\}\cap A\in\mathcal{F}_n\}}$.
For any random variable $X$ with finite expectation, and for any $n \in \mathbb{N}$, we have that
$$ \mathbb{E}(X|\mathcal{F}_\tau)\mathbf{1}_{\{\tau=n\}} =\mathbb{E}(X|\mathcal{F}_n)\mathbf{1}_{\{\tau=n\}}$$
What I'm having trouble with is the proof of this result, it goes like this in the notes :
Proof : $\forall n \in \mathbb{N}, \forall A \in \mathcal{F}_n$, we have by definition that $ {A\cap\{\tau=n\}\in\mathcal{F}_\tau} $, and we also have that ${\{\tau=n\}\in\mathcal{F}_\tau} $
From which it follows that
$$ \mathbb{E}(\mathbf{1}_AX\mathbf{1}_{\{\tau=n\}}) =\mathbb{E}(\mathbf{1}_{A\cap\{\tau=n\}}\mathbb{E}(X|\mathcal{F}_\tau)) =\mathbb{E}(\mathbf{1}_A\mathbb{E}(X\mathbf{1}_{\{\tau=n\}}|\mathcal{F}_\tau)) $$
From which we can conclude that
$$ {\mathbb{E}(X\mathbf{1}_{\{\tau=n\}}|\mathcal{F}_n) =\mathbb{E}(X\mathbf{1}_{\{\tau=n\}}|\mathcal{F}_\tau)} \qquad \blacksquare$$
What bugs me is the last step of the proof : it is true that for any $ A \in \mathcal{F}_n,{A\cap\{\tau=n\}\in\mathcal{F}_n} $, so I understand that what we have shown is
$$\mathbb{E}(\mathbf{1}_A\mathbb{E}(X\mathbf{1}_{\{\tau=n\}}|\mathcal{F}_n)) = \mathbb{E}(\mathbf{1}_A\mathbb{E}(X\mathbf{1}_{\{\tau=n\}}|\mathcal{F}_\tau)) \quad \forall A\in \mathcal{F}_n $$
But don't we need to prove this result for all $A \in \mathcal{F}$ to prove that the two conditional expectations are the same (and thus the lemma) ? Unfortunately, if we take $A$ in $\mathcal{F}$ instead of $\mathcal{F}_n$, the above proof does not hold ($\mathbf{1}_A$ is not $\mathcal{F}_n$ measurable).
If anyone could explain what I'm misunderstanding and/or why the proof is correct, I'd be grateful.
I think you need to take $A\in\mathcal{F}_{\tau}$ s.t. $A\cap \{\tau=n\}\in \mathcal{F}_n$ (using the definition of $\mathcal{F}_{\tau}$). Then $$ \mathsf{E}[\mathsf{E}[X1_{\{\tau=n\}}\mid \mathcal{F}_{\tau}]1_A]=\mathsf{E}[X1_{A\cap\{\tau=n\}}]=\mathsf{E}[\mathsf{E}[X1_{A\cap\{\tau=n\}}\mid \mathcal{F}_n]]=\mathsf{E}[\mathsf{E}[X\mid\mathcal{F}_n]1_{A\cap\{\tau=n\}}], $$ where the first equality is the definition of conditional expectations, and the second equality follows from the tower property. Consequently, $$ \mathsf{E}[X1_{\{\tau=n\}}\mid \mathcal{F}_{\tau}]=\mathsf{E}[X\mid\mathcal{F}_n]1_{\{\tau=n\}}\quad\text{a.s.} $$