Convergence in Norm Implies Convergence in Measure

Theorem
Let $\left({X, \Sigma, \mu}\right)$ be a measure space, and let $p \in \R, p \ge 1$.

Let $\left({f_n}\right)_{n \in \N}, f_n : X \to \R$ be a sequence of $p$-integrable functions.

Also, let $f: X \to \R$ be a $p$-integrable function.

Suppose that $f_n$ converges in norm to $f$ (in the $p$-norm).

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

That is:


 * $\displaystyle \operatorname{\mathcal L^{\textit p}-\!\lim\,} \limits_{n \to \infty} f_n = f \implies \operatorname{\mu-\!\lim\,} \limits_{n \to \infty} f_n = f$