Equivalence of convergence. In particular $L^p$ and in measure.

Question

Let $f_n$ be a sequence in $\mathcal{L}^2([0,1])$ that converges to $f$. Show that $f_n$ converges in measure to $f$.

I know by definition that convergence in $\mathcal{L}^2$ means $\lim_{n\to\infty}||f_n-f||_2=0 \Leftrightarrow \lim_{n\to\infty}(\int_{0}^{1}|f_n(x)-f(x)|^2dx)^{1/2}=0$ and that I need to finish the proof by showing $\lim_{n\to\infty}\{x:|f_n(x)-f(x)|>\epsilon\}=0$ But I don't see how to go about this. I tried expanding the integral but that only hows that doens't necessarily show that it is convergent in measure.

Answer 1

\begin{align*} \epsilon|\{|f_{n}-f|>\epsilon\}|\leq\int_{\{|f_{n}-f|>\epsilon\}}|f_{n}-f|dx\leq\int_{[0,1]}|f_{n}-f|dx\leq\|f_{n}-f\|_{L^{2}[0,1]}. \end{align*}