Dominated Convergence Theorem
1 Statement
Lebesgue’s Dominated Convergence Theorem (DCT) gives a sufficient condition for exchanging limits and integrals. If a sequence of functions is bounded in absolute value by an integrable function and converges pointwise (almost everywhere), then the integral of the limit is the limit of the integrals, and the convergence even holds in \(L^1\).
Let \((X,\mathcal{M},\mu)\) be a measure space and \(f_{n}:X \to [-\infty, \infty]\) measurable functions such that the pointwise limit \(f(x)=\lim_{ n \to \infty }f_{n}(x)\) exists (\(\mu\)-almost everywhere). Assume there is an integrable \(g:X \to [0,\infty]\) with \(\lvert f_{n} \rvert\leq g\) for each \(n\). Then \(f\) and every \(f_n\) are integrable, and \[ \lim_{ n \to \infty }\int_{X}f_{n}~d\mu=\int_{X}\lim_{ n \to \infty } f_{n}~d\mu=\int_{X}f~d\mu, \qquad\lim_{n\to\infty}\int_X|f_n-f|\,d\mu=0. \]
Proof. Since \(|f|\leq g\) and \(|f_n|\leq g\), all these functions are integrable. As \(g+f_{n}\geq 0\) and \(g-f_{n}\geq 0\), Fatou Lemma gives \[ \begin{align} \int (g+f)\,d\mu & \leq \liminf_{n \to \infty}\int (g+f_{n})\,d\mu, \\ \int (g-f)\,d\mu & \leq \liminf_{n \to \infty}\int (g-f_{n})\,d\mu. \end{align} \] Subtracting the finite quantity \(\int g\,d\mu\) yields \[ \int f\,d\mu\leq\liminf_{n\to\infty}\int f_n\,d\mu,\qquad \int f\,d\mu\geq\limsup_{n\to\infty}\int f_n\,d\mu, \] so the limit exists and equals \(\int f\,d\mu\). For the \(L^1\) statement apply this to \(|f_n-f|\leq 2g\), which tends to \(0\) pointwise.
2 Random Variables
Taking \(\mu=\mathbb{P}\) we obtain the version for expectations.
Let \(\{ X_{n} \}\) be random variables with \(X_n\to X\) almost surely, and suppose \(\lvert X_{n} \rvert\leq Y\) for every \(n\) with \(\mathbb{E}Y<\infty\). Then \(X\) is integrable and \[ \lim_{n \to \infty}\mathbb{E}[X_{n}]=\mathbb{E}\left[\lim_{n\to \infty} X_{n}\right]=\mathbb{E}[X]. \]
The assumption can be weakened to convergence in probability.
Assume \(X_{n}\stackrel{\mathbb{P}}{\to}X\) and \(|X_{n}|\leq Y\) with \(\mathbb{E}[Y]<\infty\). Then \(\mathbb{E}[X_{n}] \to \mathbb{E}[X]\).
Proof. Some subsequence converges to \(X\) almost surely, so \(|X|\leq Y\) a.s. and \(X\) is integrable. Suppose \(\mathbb{E}X_n\not\to\mathbb{E}X\). Then there are \(\epsilon>0\) and a subsequence \(n_k\) with \(|\mathbb{E}X_{n_k}-\mathbb{E}X|>\epsilon\). Since \(X_{n_k}\to X\) in probability, it has a further subsequence converging almost surely, along which the DCT gives \(\mathbb{E}X_{n_{k_j}}\to\mathbb{E}X\), a contradiction.