Pointwise Convergence Implies Convergence in Measure on Finite Measure Space

Theorem
Let $\struct {X, \Sigma, \mu}$ be a finite measure space.

Let $\sequence {f_n}_{n \mathop \in \N}, f_n : X \to \R$ be a sequence of measurable functions.

Also, let $f: X \to \R$ be a measurable function.

Suppose that $f$ is the pointwise limit of the $f_n$ $\mu$-almost everywhere.

Then $f_n$ converges in measure to $f$ (in $\mu$).

That is:


 * $\ds \lim_{n \mathop \to \infty} \map {f_n} x \stackrel {a.e.} {=} \map f x \implies \operatorname {\mu-\!\lim\,} \limits_{n \mathop \to \infty} f_n = f$

Proof
Since $\sequence {f_n}_{n \mathop \in \N}$ converges almost everywhere, we have:


 * $\set {x \in X : \sequence {\map {f_n} x}_{n \mathop \in \N} \text { does not converge to } \map f x}$ is a $\mu$-null set.

We aim to deduce that:


 * $\ds \lim_{n \mathop \to \infty} \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x} > \epsilon} } = 0$

for each $\epsilon > 0$.

Let $\epsilon > 0$.

For each $n \in \N$, define:


 * $A_n = \set {x \in X : \size {\map {f_n} x - \map f x} > \epsilon}$

Also define:


 * $\ds B_n = \bigcup_{k \mathop = n}^\infty A_n$

for each $n \in \N$.

Then $\sequence {B_n}_{n \mathop \in \N}$ is a decreasing sequence.

Since $\mu$ is finite, $\map \mu {B_1} < \infty$.

So, from Measure of Limit of Decreasing Sequence of Measurable Sets, we have:


 * $\ds \map \mu {\bigcap_{n \mathop = 1}^\infty B_n} = \lim_{n \mathop \to \infty} \map \mu {B_n}$

Note that if:


 * $\ds x \in \bigcap_{n \mathop = 1}^\infty B_n$

We have:


 * $\size {\map {f_n} x - \map f x} > \epsilon$ for each $n \in \N$.

So:


 * $\sequence {\map {f_n} x}_{n \mathop \in \N}$ does not converge to $\map f x$.

So:


 * $\ds \bigcap_{n \mathop = 1}^\infty B_n \subseteq \set {x \in X : \sequence {\map {f_n} x}_{n \mathop \in \N} \text { does not converge to } \map f x}$

So, from Null Sets Closed under Subset, we have:


 * $\ds \map \mu {\bigcap_{n \mathop = 1}^\infty B_n} = 0$

and so:


 * $\ds \lim_{n \mathop \to \infty} \map \mu {B_n} = 0$

That is:


 * $\ds \lim_{n \mathop \to \infty} \map \mu {\bigcup_{k \mathop = n}^\infty A_k} = 0$

From Set is Subset of Union, we have:


 * $\ds A_n \subseteq \bigcup_{k \mathop = n}^\infty A_k$

So from Measure is Monotone, we have:


 * $\ds 0 \le \map \mu {A_n} \le \map \mu {B_n}$ for each $n \in \N$.

Then, from the Squeeze Theorem for Real Sequences, we have:


 * $\ds \lim_{n \mathop \to \infty} \map \mu {A_n} = 0$

That is:


 * $\ds \lim_{n \mathop \to \infty} \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x} > \epsilon} } = 0$

Since $\epsilon > 0$ was arbitrary, we have:


 * $f_n \stackrel \mu \longrightarrow f$

Also see

 * Pointwise Convergence may not imply Convergence in Measure shows that if $\mu$ is not a finite measure, the implication may not hold