Could a limiting result (in terms of convergence) imply an equivalence relation?

Question

1. Let $(X_n)$ be a martingale and a uniformly integrable collection of random variables. Given a filtration $(\mathcal{F}_n)$, given that $\Phi\in\mathcal{F}_m$ and $X_n\rightarrow X_{\infty}$ in $\mathcal{L}^1$, knowing that $$|\mathbb{E}\{X_n1_{\Phi}\}-\mathbb{E}\{X_{\infty}1_{\Phi}\}|\leq\mathbb{E}\{|X_n-X_{\infty}|\}$$ considering that, since $X_n\rightarrow X$ in $\mathcal{L}^1$, $\lim\limits_{n\rightarrow\infty}\mathbb{E}\{|X_n-X_{\infty}|\}=0$ and therefore $\lim\limits_{n\rightarrow\infty}|\mathbb{E}\{X_n1_{\Phi}\}-\mathbb{E}\{X_{\infty}1_{\Phi}\}|=0$ ,
   is one allowed to state that $\mathbb{E}\{X_n1_{\Phi}\}=\mathbb{E}\{X_{\infty}1_{\Phi}\}$? In other terms, in general does it hold true that $$\lim\limits_{n\rightarrow\infty}|\mathbb{E}\{X_n1_{\Phi}\}-\mathbb{E}\{X_{\infty}1_{\Phi}\}|=0\Rightarrow\mathbb{E}\{X_n1_{\Phi}\}=\mathbb{E}\{X_{\infty}1_{\Phi}\}$$ If so, why?.
   
   Personally, I would end up with a limiting result in terms of relation between $\mathbb{E}\{X_n1_{\Phi}\}$ and $\mathbb{E}\{X_{\infty}1_{\Phi}\}$, NOT with an equivalence result, but I am pretty sure there is a flaw in my reasoning, since on Jacod-Protter I read the above discussed implication.
   
   

2. Assuming that $\mathbb{E}\{X_n1_{\Phi}\}=\mathbb{E}\{X_{\infty}1_{\Phi}\}$ holds true, can one state that this equality implies that $\mathbb{E}\{X_{\infty}|\mathcal{F}_n\}=X_n$ a.s.? I already know that $$\mathbb{E}\{X_{\infty}|\mathcal{F}_n\}=X_n\hspace{0.2cm} a.s. \Rightarrow \mathbb{E}\{X_n1_{\Phi}\}=\mathbb{E}\{X_{\infty}1_{\Phi}\}$$ and I know how to prove this, but I do not know whether the other direction, that is $$\mathbb{E}\{X_n1_{\Phi}\}=\mathbb{E}\{X_{\infty}1_{\Phi}\}\Rightarrow \mathbb{E}\{X_{\infty}|\mathcal{F}_n\}=X_n\hspace{0.2cm} a.s.$$ holds true as well.
   In general, this second question could be generalized as... "is 'equal expectation' equivalent to 'equal conditional expectation'? If so, one could condition on both sides of $\mathbb{E}\{X_n1_{\Phi}\}=\mathbb{E}\{X_{\infty}1_{\Phi}\}$ with respect to $\mathcal{F}_n$ and eventually get that $\mathbb{E}\{X_{\infty}|\mathcal{F}_n\}=X_n$ a.s., simply by definition of conditional expectation.
   
   However, I am pretty sure that "'equal expectation' IS NOT NECESSARILY equivalent to 'equal conditional expectation'" and I think there are good counterexamples to show this.
   Henceforth, how can one prove this other direction in a good way (which would let one eventually state that $\mathbb{E}\{X_{\infty}|\mathcal{F}_n\}=X_n$ a.s. $\iff$ $\mathbb{E}\{X_n1_{\Phi}\}=\mathbb{E}\{X_{\infty}1_{\Phi}\}$)?

Answer 1

The first implication is not true. Consider $X_n = 1/n,$ a deterministic sequence. This converges in $L^1$ to $X_\infty = 0$. Then for $\Phi = \Omega,$ we have $E[X_n1_\Phi] = 1/n\neq 0 = E[X_\infty 1_\Phi].$ The second implication is then also not true.

I suspect that you want $(X_n)$ to be a uniformly integrable martingale sequence, rather than just any sequence. Indeed, if $(X_n)_n$ is a uniformly integrable martingale sequence, we have by Doob's martingale convergence theorem that there exists $X_\infty$ such that $X_n\to X_\infty$ both a.s. and in $L^1$ and that the extended martingale property holds: $$E[X_\infty| \mathcal{F}_n] = X_n.$$ See for example Theorem 2.12 here: http://math.tkk.fi/teaching/stokanal/lecture3.pdf

This then implies the first implication you wrote. If $\Phi \in \mathcal{F}_m$ for $m\leq n$, then by definition of a filtration, we have $\mathcal{F}_m\subset \mathcal{F}_n$, so we also have $\Phi \in \mathcal{F}_n$. We then have $E[X_n1_\Phi] = E[X_\infty 1_\Phi]$, by definition of conditional expectation.