$f_n \rightarrow f$ in measure iff for every $\epsilon > 0$ there exists $N \in \Bbb N $ such that $\mu (\{\mid f_n -f \mid \geq \epsilon \})<\epsilon$ for all $n \geq N$.
For me, it seems direct to prove $f_n$ converges in measure just by definition; however, I'm stuck in the converse direction.
Ok let's attempt the converse direction: Let $\varepsilon > 0$. We want to show that:
$$ \mu (\{\mid f_n -f \mid \geq \varepsilon \}) \to 0 \; n \to \infty $$
But notice that for $\varepsilon > \tilde{\varepsilon}$ it holds that:
$$\{\mid f_n -f \mid \geq \varepsilon \}\subset \{\mid f_n -f \mid \geq \tilde{\varepsilon} \}$$
Therefore it holds that:
$$\mu(\{\mid f_n -f \mid \geq \varepsilon \})\leq \mu(\{\mid f_n -f \mid \geq \tilde{\varepsilon} \})$$
By our assumption, there also exists $N(\tilde{\varepsilon})$ such that:
$$\mu (\{\mid f_n -f \mid \geq \tilde{\varepsilon} \}) \leq \tilde{\varepsilon} \; \forall \; n\geq N(\tilde{\varepsilon})$$
Putting these together: For every $0<\tilde{\varepsilon}<\varepsilon$, there exists $N(\tilde{\varepsilon})$ such that:
$$\mu (\{\mid f_n -f \mid \geq {\varepsilon} \}) \leq \tilde{\varepsilon} \; \forall \; n\geq N(\tilde{\varepsilon})$$
Therefore it follows that $f_n \to f$ in measure.