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.