Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions

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

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

For each $n \in \N$, let $f_n : X \to \R$ be a positive simple function, such that:


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

and:


 * for each $x \in X$, the sequence $\sequence {\map {f_n} x}_{n \in \N}$ is increasing

where $\lim$ denotes a pointwise limit.

Then:


 * $\ds \int f \rd \mu = \lim_{n \mathop \to \infty} \int f_n \rd \mu$

where the integral signs denote $\mu$-integration.

Proof
Let $\EE^+$ be the space of positive simple functions.

Note that since:


 * for each $x \in X$, the sequence $\sequence {\map {f_n} x}$ is increasing

we have that:


 * $f_i \le f_j$

whenever $i \le j$.

Since $f_n \to f$, from Monotone Convergence Theorem (Real Analysis): Increasing Sequence, we further obtain:


 * $f_i \le f_j \le f$

whenever $i \le j$.

From Integral of Positive Simple Function is Monotone, we have:


 * $\ds \int f_i \rd \mu \le \int f_j \rd \mu \le \int f \rd \mu$

So the sequence:


 * $\ds \sequence {\int f_n \rd \mu}_{n \in \N}$

is increasing and bounded.

So, by Monotone Convergence Theorem (Real Analysis): Increasing Sequence, it converges with:


 * $\ds \lim_{n \mathop \to \infty} \int f_n \rd \mu \le \int f \rd \mu$

We now want to show:


 * $\ds \int f \rd \mu \le \lim_{n \mathop \to \infty} \int f_n$

at which point we will have:


 * $\ds \int f \rd \mu = \lim_{n \mathop \to \infty} \int f_n \rd \mu$

We will show that for all $g \in \EE^+$ with $g \le f$, we have:


 * $\ds \int g \rd \mu \le \lim_{n \to \infty} \int f \rd \mu$

At which point, we have:


 * $\ds \sup \set {\int g \rd \mu : g \le f, g \in \EE^+} \le \lim_{n \to \infty} \int f \rd \mu$

so that, by the definition of $\mu$-integration:


 * $\ds \int f \rd \mu \le \lim_{n \to \infty} \int f \rd \mu$

This will give us:


 * $\ds \int f \rd \mu = \lim_{n \to \infty} \int f \rd \mu$

as required.

For each $n \in \N$, define $h_n : X \to \R$ by:


 * $h_n = \min \set {g, f_n}$

From Simple Function is Measurable, we have that:


 * $g$ and $f_n$ are both $\Sigma$-measurable for each $n \in \N$.

So, from Pointwise Minimum of Measurable Functions is Measurable, we have:


 * $h_n$ is $\Sigma$-measurable for each $n \in \N$.

Also, from Pointwise Minimum of Simple Functions is Simple, we have:


 * $h_n$ is simple for each $n \in \N$.

We will show that:


 * $\ds \lim_{n \mathop \to \infty} h_n = g$

and that:


 * for each $x \in X$, the sequence $\sequence {\map {h_n} x}$ is increasing.

At which point, we will be able to apply Integral of Positive Simple Function as Limit of Integrals of Positive Simple Functions.

Let $x \in X$.

Since $f_n \to f$ and $f - g \ge 0$, there exists $N \in \N$ such that:


 * $\size {\map {f_n} x - \map f x} \le \map f x - \map g x$

for $n \ge N$.

Then:


 * $\map {f_n} x \ge \map g x$

for $n \ge N$.

So:


 * $\map {h_n} x = \map g x$

for $n \ge N$, giving:


 * $\ds \map g x = \lim_{n \mathop \to \infty} \map {h_n} x$

from Tail of Convergent Sequence.

Since $x \in X$ was arbitrary, we have:


 * $\ds g = \lim_{n \mathop \to \infty} h_n$

We move on to showing that $\sequence {\map {h_n} x}$ is increasing for each $x \in X$.

For each $x \in X$, let $N_x$ be the least $N$ such that:


 * $\map {f_n} x \ge \map g x$

If $N_x = 1$, then:


 * $\map {h_n} x = \map g x$

for each $n \in \N$, so:


 * the sequence $\sequence {\map {h_n} x}_{n \in \N}$ is constant.

Hence:


 * the sequence $\sequence {\map {h_n} x}_{n \in \N}$ is increasing.

Now suppose $N_x > 1$.

Then, for $n < N_x$, we have that:


 * $\map {h_n} x = \map {f_n} x$

So that, for $i \le j < N_x$, we have:


 * $\map {h_i} x \le \map {h_j} x < \map g x$

So:


 * $\sequence {\map {h_n} x}_{n \in \N}$ is increasing for each $x \in X$.

Applying Integral of Positive Simple Function as Limit of Integrals of Positive Simple Functions, we then have:


 * $\ds \int g \rd \mu = \lim_{n \to \infty} \int h_n \rd \mu$

We have that:


 * $h_n \le f_n$

for each $n \in \N$.

So, from Integral of Positive Measurable Function is Monotone:


 * $\ds \int h_n \rd \mu \le \int f_n \rd \mu$

for each $n \in \N$, so by Lower and Upper Bounds for Sequences: Corollary:


 * $\ds \lim_{n \to \infty} \int h_n \rd \mu \le \lim_{n \to \infty} \int f_n \rd \mu$

This gives:


 * $\ds \int g \rd \mu \le \lim_{n \to \infty} \int f_n \rd \mu$

hence the required conclusion.