Pointwise Convergence implies Convergence Almost Everywhere

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

Let $f : X \to \R$ be a $\Sigma$-measurable function.

For each $n \mathop \in \N$, let $f_n : X \to \R$ be a $\Sigma$-measurable function.

Then:


 * if $\sequence {f_n}_{n \mathop \in \N}$ converges pointwise to $f$, it converges almost everywhere to $f$.

Proof
If $\sequence {f_n}_{n \mathop \in \N}$ converges pointwise to $f$, then:


 * $\sequence {\map {f_n} x}_{n \mathop \in \N}$ converges to $\map f x$ for each $x \in X$.

So:


 * $\set {x \in X : \sequence {\map {f_n} x}_{n \mathop \in \N} \text { does not converge to } \map f x} = \O$

From Measure of Empty Set is Zero, we have:


 * $\map \mu \O = 0$

So:


 * $\map \mu {\set {x \in X : \sequence {\map {f_n} x}_{n \mathop \in \N} \text { does not converge to } \map f x} } = 0$

So:


 * $\sequence {f_n}_{n \mathop \in \N}$ converges almost everywhere to $f$.