Monotone Convergence Theorem (Measure Theory)/Corollary

== Corollary to Monotone Convergence Theorem (Measure Theory) ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $u : X \to \overline \R_{\ge 0}$ be a positive $\Sigma$-measurable function. 
Let $\sequence {u_n}_{n \mathop \in \N}$ be an sequence of positive $\Sigma$-measurable functions $u_n : X \to \overline \R_{\ge 0}$ such that: 
:$\map {u_i} x \ge \map {u_j} x$ for all $i \le j$
and:
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$
for $\mu$-almost all $x \in X$. 
Suppose also that $u_1$ is $\mu$-integrable function.
Then $u_n$ is $\mu$-integrable for each $n \in \N$ and $u$ is $\mu$-integrable with: 
:$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int u_n \rd \mu$
== Proof ==
Suppose that: 
:$\map {u_i} x \ge \map {u_j} x$ for all $i \le j$
and:
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$
for all $x \in X$. 
Then in particular we have: 
:$u_1 \ge u_n$ for each $n \in \N$. 
From Integral of Positive Measurable Function is Monotone, we have: 
:$\ds \int u_1 \rd \mu \ge \int u_n \rd \mu$
Since $u_1$ is $\mu$-integrable, we have that: 
:$u_n$ is $\mu$-integrable for each $n \in \N$. 
We also have: 
:$u_1 \ge u$
so $u$ is $\mu$-integrable.
From Integrable Function is A.E. Real-Valued, for each $k \in \N$ there exists a $\mu$-null set $N_k \subseteq X$ such that:
:whenever $\map {u_k} x = \infty$, we have $x \in N_k$.
Let: 
:$\ds N = \bigcup_{k \mathop = 1}^\infty N_k$
then: 
:$\map {u_k} x < \infty$ for each $x \in X \setminus N$ for all $k \in \N$.
Since $\sigma$-algebras are closed under countable union, we have: 
:$N$ is $\Sigma$-measurable. 
From Null Sets Closed under Countable Union, we have: 
:$N$ is $\mu$-null.
Now let $\sequence {v_n}_{n \mathop \in \N}$ be a sequence of real-valued functions $v_k : X \to \R$ with: 
:$\map {v_k} x = \map {u_1} x \times \map {\chi_{X \setminus N} } x - \map {u_k} x \times \map {\chi_{X \setminus N} } x$
We show that $\sequence {v_n}_{n \mathop \in \N}$ is an increasing sequence converging to: 
:$v = u_1 \times \chi_{X \setminus N} - u \times \chi_{X \setminus N}$
which we will verify is well-defined. 
For $x \in X \setminus N$, we then have: 
:$\map {v_k} x = \map {u_1} x - \map {u_k} x$
Since: 
:$\sequence {\map {u_k} x}_{k \mathop \in \N}$ is an decreasing sequence
we have: 
:$\sequence {\map {v_k} x}_{k \mathop \in \N}$ is a increasing sequence.
From Monotone Convergence Theorem (Real Analysis): Decreasing Sequence, we have: 
:$\map u x < \infty$ 
So we have: 
:$\ds \lim_{k \mathop \to \infty} \map {v_k} x = \map {u_1} x - \map u x = \map {u_1} x \times \map {\chi_{X \setminus N} } x - \map u x \times \map {\chi_{X \setminus N} } x$ 
for each $x \in X \setminus N$. 
Now suppose that $x \in N$. 
Then we have: 
:$\map {v_k} x = 0$
for each $k \in \N$. 
Clearly, $\sequence {\map {v_k} x}_{k \mathop \in \N}$ is increasing and converges to: 
:$0 = \map {u_1} x \times \map {\chi_{X \setminus N} } x - \map u x \times \map {\chi_{X \setminus N} } x$
So we have shown that $\sequence {v_n}_{n \mathop \in \N}$ is an increasing sequence with: 
:$v_n \to u_1 \times \chi_{X \setminus N} - u \times \chi_{X \setminus N}$
By the Monotone Convergence Theorem, we therefore have: 
:$\ds \int \paren {u_1 \times \chi_{X \setminus N} - u \times \chi_{X \setminus N} } \rd \mu = \lim_{n \mathop \to \infty} \int v_n \rd \mu$ 
That is: 
:$\ds \int \paren {u_1 \times \chi_{X \setminus N} - u \times \chi_{X \setminus N} } \rd \mu = \lim_{n \mathop \to \infty} \paren {u_1 \times \chi_{X \setminus N} - u_n \times \chi_{X \setminus N} } \rd \mu$ 
From Integral of Integrable Function is Additive: Corollary 2, we have: 
:$\ds \int \paren {u_1 \times \chi_{X \setminus N} } \rd \mu - \int \paren {u \times \chi_{X \setminus N} } \rd \mu = \int \paren {u_1 \times \chi_{X \setminus N} } \rd \mu - \lim_{n \mathop \to \infty} \int \paren {u_n \times \chi_{X \setminus N} } \rd \mu$
giving:
:$\ds \int \paren {u \times \chi_{X \setminus N} } \rd \mu = \lim_{n \mathop \to \infty} \int \paren {u_n \times \chi_{X \setminus N} } \rd \mu$
From Characteristic Function of Null Set is A.E. Equal to Zero: Corollary, we have: 
:$\chi_{X \setminus N} = 1$ $\mu$-almost everywhere.
So, from Pointwise Multiplication preserves A.E. Equality, we have: 
:$u \times \chi_{X \setminus N} = u$ $\mu$-almost everywhere
and:
:$u_n \times \chi_{X \setminus N} = u_n$ $\mu$-almost everywhere for each $n \in \N$. 
From A.E. Equal Positive Measurable Functions have Equal Integrals, we have: 
:$\ds \int \paren {u \times \chi_{X \setminus N} } \rd \mu = \int u \rd \mu$
and:
:$\ds \int \paren {u_n \times \chi_{X \setminus N} } \rd \mu = \int u_n \rd \mu$ for each $n \in \N$.
So we obtain: 
:$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int u_n \rd \mu$ 
Now suppose that:
:$\map {u_i} x \ge \map {u_j} x$ for all $i \le j$
and:
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$
for $\mu$-almost all $x \in X$.
Then, there exists a $\mu$-null set $N_1 \subseteq X$ such that whenever: 
:$\map {u_i} x < \map {u_j} x$ for some $i < j$
and:
:either $\ds \lim_{n \mathop \to \infty} \map {u_n} x$ does not exist or $\ds \map u x \ne \lim_{n \mathop \to \infty} \map {u_n} x$ 
we have $x \in N$.
For each $n \in \N$, define $v_n : X \to \overline \R_{\ge 0}$ by: 
:$\map {v_n} x = \map {u_n} x \times \map {\chi_{X \setminus N_1} } x$
for each $x \in X$. 
Also define $v : X \to \overline \R_{\ge 0}$ by:
:$\map v x = \map u x \times \map {\chi_{X \setminus N_1} } x$ 
for each $x \in X$.
Clearly, if $x \in X \setminus N$, we have: 
:$\map {v_n} x = \map {u_n} x$ for each $n$
and:
:$\map v x = \map u x$
From the definition of $N$, we have: 
:$\map {u_i} x \ge \map {u_j} x$ for all $i \le j$
and:
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$ 
for $x \in X \setminus N$.
So:
:$\map {v_i} x \ge \map {v_j} x$ for all $i \le j$
and:
:$\ds \map v x = \lim_{n \mathop \to \infty} \map {v_n} x$
If $x \in N$, we have: 
:$\map {v_n} x = 0$ for each $n \in \N$
and:
:$\map v x = 0$ 
So we have: 
:$\map {v_i} x \ge \map {v_j} x$ for $i \le j$ 
and:
:$\ds \map v x = \lim_{n \mathop \to \infty} \map {v_n} x$
for $x \in N$. 
So, we have: 
:$\map {v_i} x \ge \map {v_j} x$ for $i \le j$ 
and:
:$\ds \map v x = \lim_{n \mathop \to \infty} \map {v_n} x$
for all $x \in X$.
From our previous work, we have: 
:$\ds \int v \rd \mu = \lim_{n \mathop \to \infty} \int v_n \rd \mu$
That is: 
:$\ds \int \paren {u \times \chi_{X \setminus N} } \rd \mu = \lim_{n \mathop \to \infty} \int \paren {u_n \times \chi_{X \setminus N} } \rd \mu$ 
From Characteristic Function of Null Set is A.E. Equal to Zero: Corollary, we have: 
:$\chi_{X \setminus N} = 1$ $\mu$-almost everywhere.
So, from Pointwise Multiplication preserves A.E. Equality, we have: 
:$u \times \chi_{X \setminus N} = u$ $\mu$-almost everywhere
and:
:$u_n \times \chi_{X \setminus N} = u_n$ $\mu$-almost everywhere for each $n \in \N$.
Then from A.E. Equal Positive Measurable Functions have Equal Integrals, we have: 
:$\ds \int \paren {u \times \chi_{X \setminus N} } \rd \mu = \int u \rd \mu$
and:
:$\ds \int \paren {u_n \times \chi_{X \setminus N} } \rd \mu = \int u_n \rd \mu$ for each $n \in \N$. 
So, we get: 
:$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int u_n \rd \mu$
== Sources ==
* : $11.1 \ \text{(ii)}$
Category:Measure Theory
Category:Monotone Convergence Theorem (Measure Theory)