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
\[ E(\limsup_{\ntoinf} X_n) = 0 \not\ge \limsup_{\ntoinf} (EX_n) = \infty.\]