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I have a question regarding Fatou's Lemma and a sequence of random variables converging almost surely. Fatou's Lemma states

If $\forall n \in \mathbb{N}, \,\, X_{n} \ge 0$ and $\displaystyle X = \liminf_{n \rightarrow \infty} X_{n}$, then $\displaystyle\mathbb{E}[ \liminf_{n \rightarrow \infty}\: X_{n}] \le \liminf_{n \rightarrow \infty}\: \mathbb{E}[ X_{n}]$

Suppose we also know that $X_{n} \rightarrow X$ almost surely. How can we connect this to the requirements of Fatou's Lemma? It seems to me that the Lemma asks for pointwise convergence, a wholly different beast.

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Fatou's Lemma doesn't ask for pointwise convergence. The $\lim\inf$ always exists. –  Rasmus Sep 12 '11 at 20:13
    
In Fatou's Lemma it is sufficient to have $X = \liminf_{n\to\infty} X_n$ a. s. –  martini Sep 12 '11 at 20:15
    
In fact, convergence in probability (in measure) suffices. –  Byron Schmuland Sep 12 '11 at 23:20

1 Answer 1

up vote 2 down vote accepted

Suppose you know that $X_n \to X$ a.s. To avoid confusion, let's write $Y$ for $\liminf X_n$. Since for a convergent sequence, the limit and liminf are equal, we have $X = Y$ a.s., so $E[X] = E[Y]$, and by Fatou $E[Y] \le \liminf E[X_n]$.

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Ah! The definition and proof for Fatou's Lemma given in my text defines $Y(x) = \liminf_{n} X_{n}(x)$, so I was concerned about the points where this does not hold (even if those points form a set of measure 0). Though it might seem elementary, could you prove that if $C \subset \Omega$, $P(C) = 1$, and $f$ measurable, then $\int_{C} f(x)dP(x) = \int_{\Omega} f(x)dP(x)$? –  duckworthd Sep 13 '11 at 17:50
    
Yes. For any integrable $X$ and measurable $B$ such that $P(B)=0$, $E(Y)=0$ where $Y=X\mathbf 1_B$. Note that $X=X+Y$ almost surely implies $E(X)=E(X+Y)$. –  Did Sep 13 '11 at 19:08

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