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 Inequality Rule for Real Sequences:
- $\ds \lim_{n \mathop \to \infty} \int g_n \rd \mu \le \lim_{n \mathop \to \infty} \int f_n \rd \mu$
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 exist 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:
- $E_{n, i} = \set {x \in E_i : \map {f_n} x \ge \paren {1 - \epsilon} a_i}$
Each $f_n$ is a positive simple function.
Hence from Simple Function is Measurable:
- each $f_n$ is $\Sigma$-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:
\(\ds \lim_{n \mathop \to \infty} \int g_n \rd \mu\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \sum_{i \mathop = 1}^k \paren {1 - \epsilon} a_i \map \mu {E_{n, i} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {1 - \epsilon} \sum_{i \mathop = 1}^k a_i \paren {\lim_{n \mathop \to \infty} \map \mu {E_{n, i} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {1 - \epsilon} \sum_{i \mathop = 1}^k a_i \map \mu {E_i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {1 - \epsilon} \int f \rd \mu\) | Definition of Integral of Positive Simple Function |
as required.
$\blacksquare$