The theorem below is from Durret's Probability Theory and Example. I am struggling to follow the proof of (c). I will try to give all my thought here. By the definition of conditional expectation and (b), we have for all $A\in \mathcal{F}$
$$
\int_A X_n dP = \int_A E(X_n|\mathcal{F})dP\leq \int E(X|\mathcal{F})dP=\int_A XdP
$$
So by the squeeze theorem, $\int_A E(X_n|\mathcal{F})dP\rightarrow \int_A E(X|\mathcal{F})dP$. The tricky part is then to show this implies $ E(X_n|\mathcal{F}\rightarrow E(X|\mathcal{F})$ almost surely.
Durret's approach seems to use dominated convergence theorem and this property of integral (link to another SE question). However, I don't exactly follow his steps. Why does $Y_n\downarrow 0\implies Z_\infty$ exists? What is the random variable with finite expectation that bounds $Z_n$ (in order to use DCT)? I guess $Y_1$ works beause $E(Y_1)=E(X)-E(X_1)<\infty$ where $0\leq E(X_1\leq E(X)<\infty$.
A longer proof would be most welcome (or explain all four steps in the proof).
Proof:
Best Answer
Please let me know if the following is not clear. All equalities and inequalities of conditional expectations hold a.s. Now consider $(Z_n)_{n \in \mathbb{N}}$ which are versions of the conditional expectations $E[Y_n|\mathscr{F}]$ and some $$\omega \in \bigcap_{n \in \mathbb{N}}\{Z_n=E[Y_n|\mathscr{F}]\}\cap \{E[Y_{n+1}|\mathscr{F}]\leq E[Y_{n}|\mathscr{F}]\}\cap \{E[Y_n|\mathscr{F}]\geq 0\}$$ This implies $0\leq ...\leq Z_{n+1}(\omega)\leq Z_{n}(\omega)\leq ...\leq Z_1(\omega)$. Since the sequence $(Z_n(\omega))_{n \in \mathbb{N}}$ is monotone decreasing and bounded below by $0$, it has a limit $\lim_nZ_n(\omega)=\inf_nZ_n(\omega)$. Now, also note $Z:=\inf_nZ_n$ is a measurable function. So we conclude that $\omega \in \{Z_n\to Z\}$; since $\omega$ was arbitrary, we have the set inclusion and $$P(Z_n\not\to Z)\leq \sum_{n \in \mathbb{N}}(P(Z_n\neq E[Y_n|\mathscr{F}])+P(E[Y_{n+1}|\mathscr{F}]> E[Y_{n}|\mathscr{F}])+P(E[Y_n|\mathscr{F}]< 0))=0$$ since the sets we are measuring on the rhs of the inequality have probability zero. So $Z_n \to Z$ a.s. By definition, we also have that $0\leq Z_n\leq Z_1,\,\forall n$ a.s. so we can apply DCT.