# Can you prove Fatou's lemma for conditional expectations by that of the normal version?

I'd like to discuss proofs of Fatou's lemma for conditional expectations.

It can be proved by almost the same idea for normal version, i.e., by applying the monotone convergence theorem for conditional expectations for $$\inf_{k \geq n} X_k$$. You can review its detail by the link above toward Wikipedia.

Then, I wonder how we can prove Fatou's lemma for conditional expectations, not using the monotone convergence theorem for conditional expectations, but using Fatou's lemma of the normal version. Do you know this proof?

Letting $N = \left\{ E[\liminf_{n \to \infty} X_n | \mathcal{G}] > \liminf_{n \to \infty} E[X_n | \mathcal{G}] \right\},$ I tried to prove $$P(N) = 0$$ by applying Fatou's lemma for $$\{ E[X_n | \mathcal{G}] \}$$ and $$\{ X_n \}$$, but failed. It seemed that simple ideas didn't work well.

I consulted my textbooks for probability theory and related fields and I found only the same proof above.

Kindly please note that our definition of conditional expectations is a standard one characterized by integrals and measurableness. If we define conditional probabilities first and then introduce conditional expectations by integrals with them, Fatou's lemma for conditional expectations is proved by that of the normal version, obviously. I hope you will tell me a connection between the standard definition of conditional expectations and Fatou's lemma of the normal version.

A little late to the party but the general idea is for all $$G \in \mathcal G$$ $$\int_G E[X_n | \mathcal G]dP = \int_G X_n dP$$ I think you were on the right track with your $$N$$ set. Let $$\varepsilon>0$$. $$N = \{E[\liminf_{n \rightarrow \infty} X_n|\mathcal G] > \liminf_{n \rightarrow \infty} E[X_n | \mathcal G] + \varepsilon\} \in \mathcal G$$ $$\implies E[\liminf_{n \rightarrow \infty} X_n | \mathcal G] > E[X_n| \mathcal G] + \varepsilon \textrm{ infinitely often on }N$$ Let $$\{X_{n_k}\}$$ be the subsequence where this happens $$\implies E[\liminf_{n \rightarrow \infty} X_n|\mathcal G] > E[X_{n_k} | \mathcal G] + \varepsilon \textrm{ on }N$$ $$\implies \int_N E[\liminf_{n \rightarrow \infty} X_n|\mathcal G]dP \geq \varepsilon P(N)+ \int_N E[X_{n_k} | \mathcal G]dP$$ Now use the defining relation on both sides $$\implies \int_N \liminf_{n \rightarrow \infty} X_n dP \geq \varepsilon P(N) +\int_N X_{n_k} dP$$ And Fatou's Lemma on the LHS $$\implies \liminf_{n \rightarrow \infty }\int_N X_n dP \geq \varepsilon P(N) + \int_N X_{n_k} dP$$ Since the righthand is a subsequence, its liminf cannot be smaller, and since the inequality held for all $$k$$ we can take it. $$\implies \liminf_{n \rightarrow \infty }\int_N X_n dP \geq \varepsilon P(N) + \liminf_{k \rightarrow \infty} \int_N X_{n_k} dP \geq \varepsilon P(N) +\liminf_{n \rightarrow \infty} \int_N X_n dP$$ Now we have a squeeze, so subtract the integral from the outermost sides $$0 \geq \varepsilon P(N) \geq 0$$ $$\implies P(N) = 0$$ And since $$\varepsilon>0$$ was arbitrary $$E[\liminf_{n \rightarrow \infty} X_n |\mathcal G] \overset{a.s.}{\leq} \liminf_{n \rightarrow \infty} E[X_n | \mathcal G]$$ I think the connection between them is all in the defining relation. Fatou's Lemma essentially tells us $$\liminf_{n \rightarrow \infty}\int_\Omega f_n d\mu \geq \int_\Omega\liminf_{n \rightarrow \infty} f_n d\mu$$ Which is the same as saying for all $$\varepsilon>0$$ $$\int_\Omega f_n d\mu \not \leq \varepsilon +\int_\Omega\liminf_{n \rightarrow \infty} f_n d\mu \textrm{ infinitely often }$$