Convergence in Mean implies Convergence in Measure

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 in mean to $f$, it converges in measure to $f$.

Proof
From Pointwise Difference of Measurable Functions is Measurable:


 * $f_n - f$ is $\Sigma$-measurable for each $n \in \N$.

Let $\epsilon > 0$ be a real number.

From Markov's Inequality, we then have:


 * $\ds \map \mu {\set {x \in X : \size {\map {f_n} x - \map f x} > \epsilon} } \le \frac 1 \epsilon \int \size {f_n - f} \rd \mu$

Since $\sequence {f_n}_{n \mathop \in \N}$ converges in mean to $f$, we have:


 * $\ds \lim_{n \mathop \to \infty} \frac 1 \epsilon \int \size {f_n - f} \rd \mu = 0$

So, from the Squeeze Theorem, we have:


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

for each $\epsilon > 0$.

So:


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