Sandwiching Limsups & liminfs of expectations

Question

Why is it that if we sandwich a liminf of an expectation between two equal quantities we get that the limit exists? Can we somehow deduce the limsup from that and conclude that it's the same or am I missing something else. A book I'm reading used a rule called Fatous lemma to give the below, given that $X_n \rightarrow X$ a.s. and $\mathbb{E}|X_n| \to \mathbb{E}|X|$. \begin{equation} \mathbb{E}|X| = \lim \mathbb{E}|X_n| \ge \liminf \mathbb{E}X^+ + \liminf\mathbb{E}X_n^- \ge \mathbb{E}X^+ + \mathbb{E}X^- = \mathbb{E}|X| \end{equation}

The problem is that at this point it was concluded that $\lim \mathbb{E}X_n^+ \to \mathbb{E}X^+$ and $\lim \mathbb{E}X_n^- \to \mathbb{E}X^-$. I don't understand how can we conclude this just from the above equality. Thanks in advance.

Answer 1

Fatou's lemma is an important result in measure theory. Suppose that $(X,\mathcal{B},\mu)$ is a measure space (just think $(\Omega, \mathcal{F},\mathbb{P}))$. Fatou's lemma states that, for any measurable positive functions $f_n \geq 0$, $$\liminf_{n\rightarrow \infty} \int f_n d\mu \geq \int \liminf_{n\rightarrow\infty} f_n d\mu.$$ [One way to remember this is to think of a sequence of functions whose limit is $0$ but which all integrate to 1. Then the left-hand side above is always 1 while the right-hand side is 0.] Note that $X_n \rightarrow X$ a.s. does not guarantee $E|X_n| \rightarrow E|X|$.

Next, for fixed $\omega$, $X_n(\omega)$ is a sequence of real numbers. If $\lim_{n\rightarrow\infty} X_n(\omega)$ exists for almost all $x$, then $\lim_{n\rightarrow\infty} X_n(\omega) = \liminf_{n\rightarrow\infty} X_n(\omega)$ for all such $x$. This and $E|X_n| \rightarrow E|X|$ imply $$\lim_{n\rightarrow\infty} E X_n^\pm = \liminf_{n\rightarrow\infty} E X_n^\pm \geq E\liminf_{n\rightarrow\infty} X_n^\pm = E\lim_{n\rightarrow\infty} X_n^\pm.$$ Note that $X^\pm \geq 0$, so we may apply Fatou's lemma.

The last part you bring up, and good to notice, is that $$\liminf_{n\rightarrow\infty} EX_n^+ + \liminf_{n\rightarrow\infty}EX_n^- = EX^+ + EX^-$$ may not immediately imply equality of the two individual limits.

Write the above as $$a+b = c+d.$$ By Fatou's lemma, we have $a \geq c$ and $b \geq d$. If $a \neq c$, then $c = a-\epsilon > 0$ for $\epsilon > 0$. Hence $d = b+\epsilon > b$, a contradiction. It follows that $a = c$ and $b = d$.

Answer 2

I don't see why the mentioned equality yields the existence of the limits, but here is an alternative way to prove the (original) statement. It actually even shows $X_n \to X$ in $L^1$.

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Since $$\mathbb{E}|X_n| \to \mathbb{E}|X| \tag{1}$$ and $X \in L^1$, we have $X_n \in L^1$ for sufficiently large $n$ and

$$|X_n-X| \leq |X_n|+ |X| \in L^1.$$

This means that we can apply Fatou's lemma to the non-negative sequence $|X_n|+|X|-|X_n-X|$:

$$\begin{align*} 2 \mathbb{E}|X| &\stackrel{(1)}{=} \mathbb{E}\left( \lim_{n \to \infty} |X_n|+|X|-|X_n-X| \right) \\ &= \mathbb{E}\left( \liminf_{n \to \infty} |X_n|+|X|-|X_n-X| \right)\\ &\leq \liminf_{n \to \infty} \left( \mathbb{E}|X_n|+\mathbb{E}|X| - \mathbb{E}|X_n-X| \right) \\ &\stackrel{(1)}{=} 2 \mathbb{E}|X| - \limsup_{n \to \infty} \mathbb{E}|X_n-X|. \tag{2}\end{align*}$$

Since $|X_n-X| \geq 0$ is non-negative, we conclude

$$0 \leq \liminf_{n \to \infty} \mathbb{E}|X_n-X| \leq \limsup_{n \to \infty} \mathbb{E}|X_n-X| \stackrel{(2)}{\leq} 0.$$

Hence,

$$\lim_{n \to \infty} \mathbb{E}|X_n-X| = 0,$$

i.e. $X_n \to X$ in $L^1$. Now note that the mapping $f(x) := x \mapsto \max\{0,x\}$ is contractive (i.e. $|f(x)-f(y)| \leq |x-y|$). We therefore find

$$|\mathbb{E}(X_n^+-X^+)| = |\mathbb{E}(f(X_n)-f(X))| \leq \mathbb{E}|f(X_n)-f(X)| \leq \mathbb{E}|X_n-X| \to 0.$$

Similarly, we get $\mathbb{E}X_n^- \to \mathbb{E}X^-$.