User:Caliburn/s/mt/1
Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space, where $\mu$ is a finite measure.
Let $f : X \to \R$ be a $\Sigma$-measurable function.
For each $n \in \N$, let $f_n : X \to \R$ be a $\Sigma$-measurable function.
Suppose that $\sequence {f_n}_{n \in \N}$ converges almost everywhere to $f$.
Then $\sequence {f_n}_{n \in \N}$ converges in measure to $f$.
Proof
Let $\epsilon$ be a positive real number.
For each $n \in \N$, define:
- $A_n = \set {x \in X : \size {\map {f_n} x - \map f x} > \epsilon}$
and:
- $\ds B_n = \bigcup_{k \mathop = n}^\infty A_k$
That is:
- $B_n = \set {x \in X : \size {\map {f_n} x - \map f x} > \epsilon \text { for some } k \ge n}$
From Pointwise Difference of Measurable Functions is Measurable, we have:
- $f_n - f$ is $\Sigma$-measurable for each $n \in \N$.
Then, from Absolute Value of Measurable Function is Measurable, we have:
- $\size {f_n - f}$ is $\Sigma$-measurable for each $n \in \N$.
So, from Characterization of Measurable Functions:
- $A_n$ is $\Sigma$-measurable for each $n \in \N$.
Since $\sigma$-Algebras are closed under countable intersection, we have:
- $B_n$ is $\Sigma$-measurable for each $n \in \N$.
We have:
\(\ds \bigcap_{n \mathop = 1}^\infty B_n\) | \(=\) | \(\ds \bigcap_{n \mathop = 1}^\infty \set {x \in X : \size {\map {f_n} x - \map f x} > \epsilon \text { for some } k \ge n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {x \in X : \text {for all } n \in \N \text { there exists } k \ge n \text { with } \size {\map {f_n} x - \map f x} > \epsilon}\) |
Clearly if $x \in X$, then:
- $\sequence {\map {f_n} x}_{n \in \N}$ does not converge.
So:
- $\ds \bigcap_{n \mathop = 1}^\infty B_n \subseteq \set {x \in X : \map {f_n} x \text { does not converge to } \map f x}$
Since $f$ converges almost everywhere, we have:
- $\map \mu {\set {x \in X : \map {f_n} x \text { does not converge to } \map f x} } = 0$
So, from Null Sets Closed under Subset:
- $\ds \map \mu {\bigcap_{n \mathop = 1}^\infty B_n} = 0$
Since $\mu$ is finite, we have that:
- $\map \mu {B_n}$ is finite for each $n \in \N$.
Clearly also:
- $\ds \bigcup_{k \mathop = n + 1}^\infty A_k \subseteq \bigcup_{k \mathop = n}^\infty A_k$
so that:
- $B_{n + 1} \subseteq B_n$
So, we can apply Measure of Limit of Decreasing Sequence of Sets to obtain:
- $\ds \map \mu {\bigcap_{n \mathop = 1}^\infty B_n} = \lim_{n \mathop \to \infty} \map \mu {B_n}$
Then:
- $\ds \lim_{n \mathop \to \infty} \map \mu {B_n} = 0$
Since:
- $\ds A_n \subseteq B_n$
We have, from Measure is Monotone:
- $\map \mu {A_n} \le \map \mu {B_n}$
for each $n \in \N$.
So, from Inequality Rule for Real Sequences:
- $\ds \lim_{n \mathop \to \infty} \map \mu {A_n} \le \lim_{n \mathop \to \infty} \map \mu {B_n} = 0$
So:
- $\ds \lim_{n \mathop \to \infty} \map \mu {A_n} = 0$
So, by the definition of $A_n$, we have:
- $\ds \lim_{n \mathop \to \infty} \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x} > \epsilon} } = 0$
So:
- $f$ converges in measure.