Integral of Positive Measurable Function is Positive Homogeneous

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Theorem

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

Let $f : X \to \overline \R$ be a positive $\Sigma$-measurable function.

Let $\lambda \in \overline \R$ be an extended real number with $\lambda \ge 0$.


Then:

$\ds \int \lambda f \rd \mu = \lambda \int f \rd \mu$

where:

$\lambda f$ is the pointwise $\lambda$-multiple of $f$
The integral sign denotes $\mu$-integration


This can be summarized by saying that $\ds \int \cdot \rd \mu$ is positive homogeneous.


Corollary

Let $A \in \Sigma$.


Then:

$\ds \int_A \lambda f \rd \mu = \lambda \int_A f \rd \mu$

where:

$\lambda f$ is the pointwise $\lambda$-multiple of $f$
the integral sign denotes $\mu$-integration over $A$.

This can be summarized by saying that $\ds \int_A \cdot \rd \mu$ is positive homogeneous.


Proof

Suppose that $\lambda < \infty$.

From Measurable Function is Pointwise Limit of Simple Functions, there exists an increasing sequence $\sequence {f_n}_{n \mathop \in \N}$ of positive simple functions such that:

$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$

From the Multiple Rule for Real Sequences, we have:

$\ds \lambda \map f x = \lim_{n \mathop \to \infty} \paren {\lambda \map {f_n} x}$

for each $x \in X$.

We then have:

\(\ds \int \lambda f \rd \mu\) \(=\) \(\ds \lim_{n \mathop \to \infty} \int \lambda f_n \rd \mu\) Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions
\(\ds \) \(=\) \(\ds \lambda \lim_{n \mathop \to \infty} \int f_n \rd \mu\) Integral of Positive Simple Function is Positive Homogeneous
\(\ds \) \(=\) \(\ds \lambda \int f \rd \mu\) Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions

so we get the demand in the case $\lambda < \infty$.


Now suppose that $\lambda = \infty$.

We can write:

$\ds \lambda f = \lim_{k \mathop \to \infty} k f$

Since $f \ge 0$, the sequence $\sequence {k f}_{k \mathop \in \N}$ is increasing, we have:

$\ds \int \lambda f \rd \mu = \lim_{k \mathop \to \infty} \int k f \rd \mu$

from the monotone convergence theorem.

From our earlier work, we have:

$\ds \int k f \rd \mu = k \int f \rd \mu$

so that:

\(\ds \int \lambda f \rd \mu\) \(=\) \(\ds \lim_{k \to \infty} k \int f \rd \mu\)
\(\ds \) \(=\) \(\ds \paren {\lim_{k \to \infty} k} \int f \rd \mu\)
\(\ds \) \(=\) \(\ds \lambda \int f \rd \mu\)

giving the demand in the case $\lambda = \infty$.

$\blacksquare$


Also see


Sources

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