Pointwise Convergence Implies Convergence in Measure on Finite Measure Space

Theorem
Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $\left({f_n}\right)_{n \in \N}, f_n : X \to \R$ 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:


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