1 Dominated Convergence Theorem
Lebesgue’s Dominated Convergence Theorem gives a sufficient condition for when the limits and integrals of a sequence of functions can be exchanged. More technically it says that if a sequence of functions is bounded in absolute value by an integrable function and is almost everywhere point wise convergent to a function then the sequence converges in \(L_{1}\) to its point wise limit, and in particular the integral of the limit is the limit of the integrals.
1.1 Dominated Convergence Theorem (Lebesgue Integrals)
Suppose \(f_{n}:\mathbb{R} \to [-\infty, \infty]\) are (Lebesgue) measurable functions such that the pointwise limit \(f(x)=\lim_{ n \to \infty }f_{n}(x)\) exists. Assume there is an integrable \(g:\mathbb{R} \to [0,\infty]\) with \(\lvert f_{n}(x) \rvert\leq g(x)\) for each \(x \in \mathbb{R}\). Then \(f\) is integrable as is \(f_{n}\) for each \(n\), and \[ \lim_{ n \to \infty }\int_{\mathbb{R}}^{}{f_{n}}~d{\mu}=\int_{\mathbb{R}}^{}{\lim_{ n \to \infty } f_{n}}~d{\mu}=\int_{\mathbb{R}}^{}{f}~d{\mu}. \]
PROOF:
1.2 Dominated Convergence Theorem (Random Variables)
For sequence of random variables \(\{ X_{n} \}\) such that the pointwise limit exists i.e. \(X(\omega)=\lim_{n\to \infty}X_{n}(\omega),~\forall \omega\), and is dominated by an integrable random variable \(Y\) fpr every \(n \geq 1\) i.e. \(\lvert X_{n}(\omega) \rvert\leq Y(\omega)\) and \(\mathbb{E}Y<\infty\), then \(X\) is integrable and \[ \lim_{n \to \infty}\mathbb{E}[X_{n}]=\mathbb{E}\left[\lim_{n\to \infty} X_{n}\right]=\mathbb{E}[X]. \]
Proof: Since \(-Y\leq X_{n}\leq Y\) we have that \(X_{n}+Y\geq 0\) and \(-X_{n}+Y \geq 0\) almost everywhere, and hence applying Fatou's Lemma we have that \[ \begin{align} \mathbb{E}[X+Y] & \leq \liminf_{n \to \infty}{\mathbb{E}[X_{n}+Y]} \\ \mathbb{E}[-X+Y] & \leq \liminf_{n \to \infty}{\mathbb{E}[X_{n}+Y]}. \end{align} \] Since \(Y\) is integrable, we have that \[ \begin{align} \mathbb{E}[X] & \leq \liminf_{n\to \infty}{\mathbb{E}[X_{n}]} \\ \mathbb{E}[-X] & \leq \liminf_{n \to \infty}{\mathbb{E}[-X_{n}]}\implies \mathbb{E}[X]\geq \limsup_{n \to \infty}{\mathbb{E}[X_{n}]}, \end{align} \] and since \(\liminf_{n \to \infty}{\mathbb{E}[X_{n}]}=\limsup_{n \to \infty}{\mathbb{E}[X_{n}]}\) we obtain equality. \(\square\)