Dominated Convergence Theorem

In measure theory, the dominated convergence theorem is a cornerstone of Lebesgue integration. It can be viewed as a culmination of all efforts, and is a general statement about the interplay between limits and integrals.

Theorem Statement

Consider the measure space $(X,{\mathcal {M}},\lambda )$. Suppose $\{f_{n}\}$ is a sequence in $L^{1}(\lambda )$ such that

1. $f_{n}\to f$ a.e
2. there exists $g\in L^{1}(\lambda )$ such that $|f_{n}|\leq g$ a.e. for all $n\in \mathbb {N}$

Then $f\in L^{1}(\lambda )$ and $\int f=\lim _{n\to \infty }\int f_{n}$.

Proof of Theorem

$f$ is a measurable function in the sense that it is a.e. equal to a measurable function, since it is the limit of $f_{n}$ except on a null set. Also $|f|\leq g$ a.e., so $f\in L^{1}(\lambda )$.

Now we have $g+f_{n}\geq 0$ a.e. and $g-f_{n}\geq 0$ a.e. to which we may apply Fatou's lemma to obtain

$\int g+\int f=\int \lim _{n\to \infty }(g+f_{n})\leq \liminf _{n\to \infty }\int (g+f_{n})=\int g+\liminf _{n\to \infty }\int f_{n}$,

where the equalities follow from linearity of the integral and the inequality follows from Fatou's lemma. We similarly obtain

$\int g-\int f=\int \lim _{n\to \infty }(g-f_{n})\leq \liminf _{n\to \infty }\int (g-f_{n})=\int g-\limsup _{n\to \infty }\int f_{n}$.

Since $\int g<+\infty$, these imply

$\limsup _{n\to \infty }\int f_{n}\leq \int f\leq \liminf _{n\to \infty }\int f_{n}$ from which the result follows.

Applications of Theorem

1. Suppose we want to compute $\lim _{n\to \infty }\int _{[0,1]}{\frac {1+nx^{2}}{(1+x^{2})^{n}}}$. Denote the integrand $f_{n}$ and see that $|f_{n}|\leq 1$ for all $n\in \mathbb {N}$ and $1_{[0, 1]} \in L^1(\lambda)$. Note we only consider the constant function $1$ on $[0, 1]$. Applying the dominated convergence theorem, this allows us the move the limit inside the integral and compute it as usual.

1. Using the theorem, we know there does not exist a dominating function for the sequence $f_n$ defined by $f_n(x) = n1_{[0, \frac{1}{n}]}$ because $f_n \to 0$ pointwise everywhere and $\lim _{n\to \infty }\int f_{n}=1\neq 0=\int \lim _{n\to \infty }f_{n}$.