Monotone Convergence Theorem for Positive Simple Functions

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

Let $f: X \to \R$ be a positive simple function.

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

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 $\sequence {\map {f_n} x}_{n \mathop \in \N}$ 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
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:


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

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:


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

Let $0 < \epsilon < 1$.

We will construct a non-increasing sequence of positive simple functions $\sequence {g_n}$ such that:


 * $g_n \le f_n$

and:


 * $\ds \lim_{n \mathop \to \infty} \int g_n \rd \mu = \paren {1 - \epsilon} \int f \rd \mu$

From Integral of Positive Simple Function is Monotone:


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

So, from Lower and Upper Bounds for Sequences: Corollary, we will have:


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

giving:


 * $\ds \paren {1 - \epsilon} \int f \rd \mu \le \lim_{n \mathop \to \infty} \int f_n \rd \mu$

Since $\epsilon$ is arbitrary, we will then have:


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

giving the result.

From Simple Function has Standard Representation:


 * there exists disjoint $\Sigma$-measurable sets $E_1, E_2, \ldots, E_k$ and non-negative real numbers $a_1, a_2, \ldots, a_k$ such that:


 * $\ds \map f x = \sum_{i \mathop = 1}^k a_i \map {\chi_{E_i} } x$

For each $n \in \N$ and $i \in \N$, define:


 * $E_{n, i} = \set {x \in E_i : \map {f_n} x \ge \paren {1 - \epsilon} a_i}$

Since each $f_n$ is a positive simple function, we have that each $f_n$ is $\Sigma$-measurable by Simple Function is Measurable.

So, from Characterization of Measurable Functions:


 * $E_{n, i}$ is $\Sigma$-measurable for each $n \in \N$ and $i \in \N$.

Since $f_n \le f_{n + 1}$, we have:


 * $\set {x \in E_i : \map {f_n} x \ge \paren {1 - \epsilon} a_i} \subseteq \set {x \in E_i : \map {f_{n + 1} } x \ge \paren {1 - \epsilon} a_i}$

That is:


 * $E_{n, i} \subseteq E_{\paren {n + 1}, i}$

So:


 * for each $i \in \N$, the sequence $\sequence {E_{n, i} }_{n \mathop \in \N}$ is increasing.

We now show that, for each $i \in \N$:


 * $\ds E_i = \bigcup_{n \mathop = 1}^\infty E_{n, i}$

Since:


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

for each $x \in X$, there exists $N$ such that:


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

for $n \ge N$.

In particular, for $x \in E_i$, we have:


 * $\map {f_N} x \ge \paren {1 - \epsilon} \map f x = \paren {1 - \epsilon} a_i$

So:


 * $x \in E_{N, i}$

So:


 * $\ds E_i \subseteq \bigcup_{n \mathop = 1}^\infty E_{n, i}$

By construction we have $E_{n, i} \subseteq E_i$ for each $n \in \N$, so we obtain:


 * $\ds E_i = \bigcup_{n \mathop = 1}^\infty E_{n, i}$

as required.

We have that:


 * for each $i \in \N$, $\sequence {E_{n, i} }_{n \mathop \in \N}$ is an increasing sequence of sets with $E_{n, i} \uparrow E_i$.

So, from Measure of Limit of Increasing Sequence of Measurable Sets, we have:


 * $\ds \map \mu {E_i} = \lim_{n \mathop \to \infty} \map \mu {E_{n, i} }$

Now note that since each $E_i$ is disjoint, and:


 * $E_{n, i} \subseteq E_i$

for each $n \in \N$, we must have:


 * $E_{n, i} \cap E_{n, j} = \O$ for all $n, i, j \in \N$ with $i \ne j$.

Since the sets $E_{n, 1}, E_{n, 2}, \ldots, E_{n, k}$ are all disjoint and $\Sigma$-measurable, define a positive simple function $g_n : X \to \R$ by:


 * $\ds \map {g_n} x = \sum_{i \mathop = 1}^k \paren {1 - \epsilon} a_i \map {\chi_{E_{n, i} } } x$

for each $x \in X$ and $n \in \N$.

Then from the definition of the $\mu$-integral, we have:


 * $\ds \int g_n \rd \mu = \sum_{i \mathop = 1}^k \paren {1 - \epsilon} a_i \map \mu {E_{n, i} }$

So:

as required.