Limits and Integrals

Under certain circumstances we are allowed to interchange expectation with limits.

Theorem 1. Monotone Convergence Theorem (MCT). If
$$0\leq X_n \uparrow X$$ then
$$0\leq E(X_n) \uparrow E(X)$$
Corollary 1.  Series Version of MCT.  If $X_n \ge 0$ are non-negative random variables for $n\ge1$, then
$$E(\sum_{n=1}^\infty X_n)= \sum_{n=1}^\infty E(X_n)$$
so that the expectation and infinite sum can be interchanged

Theorem 2. Fatou Lemma. If $X_n \ge 0$, then
$$E(\underset{n\rightarrow \infty}{\text{lim inf}} X_n ) \leq \underset{n\rightarrow \infty}{\text{lim inf}}E(X_n)$$
More generally, if there exists $Z\in L_1$ and $X_n\ge Z$, then
$$E(\underset{n\rightarrow \infty}{\text{lim inf}} X_n ) \leq \underset{n\rightarrow \infty}{\text{lim inf}}E(X_n)$$
Corollary 2. More Fatou.  If $0 \leq X_n \leq Z$ where $Z\in L_1$, then
$$E(\underset{n\rightarrow \infty}{\text{lim sup}} X_n ) \ge \underset{n\rightarrow \infty}{\text{lim sup}}E(X_n)$$
Theorem 3. Dominated Convergence Theorem (DCT).  If
$$X_n \rightarrow X$$ and there exists a dominating random variable $Z\in L_1$ such that
$$|X_n| \leq Z$$ then
$$E(X_n)\rightarrow E(X) \; \text{and} \; E|X_n-X|\rightarrow 0.$$
$\newcommand{\scriptB}{\mathcal{B}} \newcommand{\scriptP}{\mathcal{P}} \newcommand{\vecX}{\mathbf{X}} \newcommand{\vecx}{\mathbf{x}} \newcommand{\reals}{\mathbb{R}} \newcommand{\cplxs}{\mathbb{C}} \newcommand{\rationals}{\mathbb{Q}} \newcommand{\naturals}{\mathbb{N}} \newcommand{\integers}{\mathbb{Z}} \newcommand{\ntoinf}{n\rightarrow\infty} \newcommand{\mtoinf}{m\rightarrow\infty} \newcommand{\tendsto}{\rightarrow}$
Example of when interchanging limits and integrals without the dominating condition.  (When something very nasty happens on a small set and the degree of nastiness overpowers the degree of smallness).

Let
$$(\Omega, \scriptB, P) = ([0,1], \scriptB([0,1]), \lambda)$$ $\lambda$ the Lebesgue measure. Define
$$X_n = n^2 1_{(0,1/n)}.$$ For any $\omega \in [0,1]$,
$$1_{(0,1/n)}(w) \tendsto 0,$$ so
$$X_n \tendsto 0.$$ However
$$E(X_n) = n^2 \cdot \frac{1}{n} = n \tendsto \infty,$$ so
$$E(\liminf_{\ntoinf} X_n) = 0 \le \liminf_{\ntoinf} (EX_n) = \infty$$ and
$$E(\limsup_{\ntoinf} X_n) = 0 \not\ge \limsup_{\ntoinf} (EX_n) = \infty.$$