Integral of Series of Positive Measurable Functions

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Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$, $f_n: X \to \overline \R$ be a sequence of positive measurable functions.

Let $\ds \sum_{n \mathop \in \N} f_n: X \to \overline \R$ be the pointwise series of the $f_n$.


Then:

$\ds \int \sum_{n \mathop \in \N} f_n \rd \mu = \sum_{n \mathop \in \N} \int f_n \rd \mu$

where the integral sign denotes $\mu$-integration.


Proof

Define the sequence $\sequence {g_N}_{n \mathop \in \N}$ of functions $g_N : X \to \overline \R$ by:

$\ds \map {g_N} x = \sum_{n \mathop = 1}^N \map {f_n} x$

Since $f_n \ge 0$ for each $n$, we have:

\(\ds \map {g_{N + 1} } x\) \(=\) \(\ds \sum_{n \mathop = 1}^{N + 1} \map {f_n} x\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^N \map {f_n} x + \map {f_{N + 1} } x\)
\(\ds \) \(\ge\) \(\ds \sum_{n \mathop = 1}^N \map {f_n} x\)
\(\ds \) \(=\) \(\ds \map {g_N} x\)

So $\sequence {g_N}_{N \mathop \in \N}$ is increasing.

From Pointwise Sum of Measurable Functions is Measurable: General Result, we have that:

$g_N$ is $\Sigma$-measurable for each $N$.

So, from the monotone convergence theorem, we have:

$\ds \lim_{N \mathop \to \infty} \int g_N \rd \mu = \int \paren {\lim_{N \mathop \to \infty} g_N} \rd \mu$

From the definition of infinite series, we have:

$\ds \lim_{N \mathop \to \infty} g_N = \sum_{n \mathop = 1}^\infty f_n$

So, we have:

$\ds \lim_{N \mathop \to \infty} \int g_N \rd \mu = \int \paren {\sum_{n \mathop = 1}^\infty f_n} \rd \mu$

Manipulating the left hand side, we have:

\(\ds \lim_{N \mathop \to \infty} \int g_N \rd \mu\) \(=\) \(\ds \lim_{N \mathop \to \infty} \int \paren {\sum_{n \mathop = 1}^N f_n} \rd \mu\)
\(\ds \) \(=\) \(\ds \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N \paren {\int f_n \rd \mu}\) Integral of Positive Measurable Function is Additive
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {\int f_n \rd \mu}\) Definition of Real Series

So:

$\ds \sum_{n \mathop = 1}^\infty \paren {\int f_n \rd \mu} = \int \paren {\sum_{n \mathop = 1}^\infty f_n} \rd \mu$

$\blacksquare$


Sources

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