convergence in $L^p$ implies convergence in measure
Solution 1:
The case $p = \infty$ is the simplest since $\|f_n-f\|_{\infty} < \epsilon$ means that $|f_n-f|$ is less than $\epsilon$ almost everywhere, and therefore $\mu\big(|f_n-f|\ge\epsilon\big) = 0$.
If $1\le p < \infty$, you can use Tchebychev's inequality: $$ \mu\big(|f_n-f|\ge\epsilon\big) = \mu\big(|f_n-f|^p\ge\epsilon^p\big) \le \frac{1}{\epsilon^p}\int|f_n-f|^p\,d\mu = \frac{1}{\epsilon^p}\|f_n-f\|_p^p, $$ where the last expression can be made arbitrarily small by taking $n$ sufficiently large.