Pointwise Convergence Implies Convergence in Measure on Finite Measure Space

Theorem
Let $\struct {X, \Sigma, \mu}$ be a 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:


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