Monotone Convergence Theorem (Measure Theory)

From ProofWiki
Jump to navigation Jump to search

This proof is about Monotone Convergence Theorem in the context of Measure Theory. For other uses, see Monotone Convergence Theorem.


September 2022:  It has been suggested that this page or section be merged into Beppo Levi's Theorem .
In particular: The same up to null set
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Mergeto}} from the code.



Theorem

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:

$\map {u_i} x \le \map {u_j} x$ for all $i \le j$

and:

$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$

hold for $\mu$-almost all $x \in X$.


Then:

$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int u_n \rd \mu$


Corollary

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

First suppose that:

$\map {u_i} x \le \map {u_j} x$ for all $i \le j$

and:

$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$

for all $x \in X$.

From Integral of Positive Measurable Function is Monotone, we have that:

$\ds \int u_i \rd \mu \le \int u_j \rd \mu$ for all $i \le j$.

From Monotone Convergence Theorem (Real Analysis): Increasing Sequence, we have:

$u_i \le u$ for each $i$.

So, applying Integral of Positive Measurable Function is Monotone again we have:

$\ds \int u_i \rd \mu \le \int u \rd \mu$ for each $i$.

So:

the sequence $\ds \sequence {\int u_i \rd \mu}_{i \in \N}$ is increasing.

So, from Monotone Convergence Theorem (Real Analysis): Increasing Sequence and Unbounded Monotone Sequence Diverges to Infinity: Increasing, we have:

$\ds \sequence {\int u_i \rd \mu}_{i \in \N}$ converges, possibly to $+\infty$.

From Lower and Upper Bounds for Sequences, we also have:

$\ds \lim_{n \mathop \to \infty} \int u_n \rd \mu \le \int u \rd \mu$

We now aim to prove:

$\ds \int u \rd \mu \le \lim_{n \mathop \to \infty} \int u_n \rd \mu$

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

$\ds u_n = \lim_{k \mathop \to \infty} u_{n, k}$

From Monotone Convergence Theorem (Real Analysis): Increasing Sequence, this is equivalent to:

$\ds u_n = \sup_{k \in \N} u_{n, k}$

Let:

$\ds g_n = \max \set {u_{1, n}, u_{2, n}, \ldots, u_{n, n} }$

for each $n$.

From Pointwise Maximum of Simple Functions is Simple, $g_n$ is a positive simple function for each $n \in \N$.


With a view to apply Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions, we show that:

$\sequence {g_n}_{n \mathop \in \N}$ is increasing

and:

$\ds \map u x = \lim_{n \to \infty} \map {g_n} x$

for each $x \in X$.

We also show that:

$\map {g_n} x \le \map {u_n} x$

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

Lemma

$(1) \quad$ $\sequence {g_n}_{n \mathop \in \N}$ is increasing
$(2) \quad$ $\ds \map u x = \lim_{n \mathop \to \infty} \map {g_n} x$
$(3) \quad$ $g_n \le u_n$ for each $n \in \N$.

$\Box$


So, from Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions, we have:

$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int g_n \rd \mu$

Since:

$g_n \le u_n$ for each $n \in \N$

we have:

$\ds \int g_n \rd \mu \le \int u_n \rd \mu$ for each $n \in \N$

from Integral of Positive Measurable Function is Monotone.

So, from Inequality Rule for Real Sequences:

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

So, we have:

$\ds \int u \rd \mu \le \lim_{n \mathop \to \infty} \int u_n \rd \mu$

We therefore obtain:

$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int u_n \rd \mu$

as required.


Now suppose that:

$\map {u_i} x \le \map {u_j} x$ for all $i \le j$

and:

$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$

for $\mu$-almost all $x \in X$.

That is, there exists a $\mu$-null set $N \subseteq X$ such that whenever $x \in X$ has:

$\map {u_i} x > \map {u_j} x$ 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:

$\map {v_n} x = \map {u_n} x \times \map {\chi_{X \setminus N} } x$

for each $x \in X$.

Also define $v : X \to \overline \R_{\ge 0}$ by:

$\map v x = \map u x \times \map {\chi_{X \setminus N} } x$

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:

$\map {u_i} x \le \map {u_j} x$ 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:

$\map {v_i} x \le \map {v_j} x$ 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:

$\map {v_i} x \le \map {v_j} x$ 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:

$\map {v_i} x \le \map {v_j} x$ for $i \le j$

and:

$\ds \map v x = \lim_{n \mathop \to \infty} \map {v_n} x$

for all $x \in X$.

So, from our previous work, we have:

$\ds \int v \rd \mu = \lim_{n \mathop \to \infty} \int v_n \rd \mu$

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 $n \in \N$.

So, we have:

$u = v$ $\mu$-almost everywhere

and:

$u_n = v_n$ $\mu$-almost everywhere for each $n \in \N$.

From A.E. Equal Positive Measurable Functions have Equal Integrals, we therefore have:

$\ds \int u \rd \mu = \int v \rd \mu$

and:

$\ds \int u_n \rd \mu = \int v_n \rd \mu$ for each $n \in \N$

giving:

$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int u_n \rd \mu$

hence the demand.

$\blacksquare$


Proof 2

Let $Y \subseteq X$ be the set of $x \in X$ such that:

$\map {u_i} x \le \map {u_j} x$ for all $i \le j$

and:

$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$

Then by hypothesis:

$\map \mu {X \setminus Y} = 0$

We define an increasing sequence $\sequence {f_n}_{n \mathop \in \N}$ by:

$\map {f_n} x = \begin{cases}

\map {u_n} x & x \in Y \\ 0 & x \not \in Y \end{cases}$

Then:

\(\ds \lim_{n \mathop \to \infty} \int u_n \rd \mu\) \(=\) \(\ds \lim_{n \mathop \to \infty} \int_Y u_n \rd \mu\) Measurable Function Zero A.E. iff Absolute Value has Zero Integral
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \int f_n \rd \mu\)
\(\ds \) \(=\) \(\ds \int \sup_{n \in \N} f_n \rd \mu\) Beppo Levi's Theorem
\(\ds \) \(=\) \(\ds \int_Y \sup_{n \in \N} u_n \rd \mu\) Measurable Function Zero A.E. iff Absolute Value has Zero Integral
\(\ds \) \(=\) \(\ds \int_Y u \rd \mu\)
\(\ds \) \(=\) \(\ds \int u \rd \mu\) Measurable Function Zero A.E. iff Absolute Value has Zero Integral

$\blacksquare$

Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Monotone_Convergence_Theorem_(Measure_Theory)&oldid=592180"