Convergence Theorems

For all statements on this page, assume that $(X, A, μ)$ is a measure space.

Monotone Convergence Theorem

For each $n \in N$ let $f_{n} : X \to \overline{R}$ be a measurable function. If $0 \leq f_{n} \leq f_{n + 1}$ almost everywhere, then
$\int_{X} n \to \infty lim f_{n} d μ = n \to \infty lim \int_{X} f_{n} d μ .$

Note that the pointwise limit $lim_{n \to \infty} f_{n}$ always exists and is measurable by this proposition.

Fatou’s Lemma

For each $n \in N$ let $f_{n} : X \to \overline{R}$ be a nonnegative measurable function. Then
$\int_{X} n \to \infty lim inf f_{n} d μ \leq n \to \infty lim inf \int_{X} f_{n} d μ .$

In the following proof we omit $X$ and $d μ$ for visual clarity.

Proof By definition, we have $lim inf_{n \to \infty} f_{n} = lim_{n \to \infty} g_{n}$ , where $g_{n} = in f_{k \geq n} f_{k}$ . Now $(g_{n})$ is a monotonic sequence of nonnegative measurable functions. By the Monotone Convergence Theorem

\int n \to \infty lim inf f_{n} = n \to \infty lim \int g_{n} .

For all $k \geq n$ one has $g_{n} \leq f_{k}$ , hence $\int g_{n} \leq \int f_{k}$ by the monotonicity of the integral. This implies

\int g_{n} \leq k \geq n in f \int f_{k}

for all $n \in N$ . In the limit $n \to \infty$ we obtain

n \to \infty lim \int g_{n} \leq n \to \infty lim inf \int f_{n}

thereby completing the proof. $□$

Dominated Convergence Theorem

Let $(X, A, μ)$ be a measure space. For each $n \in N$ let $f_{n} : X \to \overline{R}$ (or $C$ ) be a measurable function. Suppose that the pointwise limit $f = lim_{n \to \infty} f_{n}$ exists almost everywhere. Suppose further that there exists an integrable function $g : X \to \overline{R}$ such that $∣ f_{n} ∣ \leq g$ almost everywhere for all $n \in N$ . Then the functions $f_{n}$ and $f$ are all integrable, and
$n \to \infty lim \int_{X} f_{n} d μ = \int_{X} f d μ .$

Proof TODO $□$