Real Analysis, Folland Proposition 2.29 Modes of Convergence Background Information:
$f_n\rightarrow f$ in $L^1$  $\Leftrightarrow$ $\forall\epsilon > 0,\exists N$ $\forall n\geq N$ $\int |f_n - f| < \epsilon$
A sequence $\{f_n\}$ of measurable complex-valued function on $(X,M,\mu)$ converges in measure to $f$ if for every $\epsilon > 0$, $$\mu\left\{x:|f_n(x) - f(x)|\geq \epsilon\}\right) \rightarrow 0 \ \ \text{as} \ \ n\rightarrow \infty$$
Question:

Proposition 2.29 - If $f_n\rightarrow f$ in $L^1$, then $f_n\rightarrow f$ in measure.

Attempted proof - Let $E_{n,\epsilon} = \{x:|f_n(x) - f(x)|\geq \epsilon\}$. Then, since $f_n\rightarrow f$ in $L^1$ we have for all $\epsilon > 0$ there exists an $N\in\mathbb{N}$ such that $$\int_{E_{n,\epsilon}}|f_n - f| \leq \int |f_n - f| < \epsilon \ \ \forall n\geq N$$
We observe that $$\int |f_n - f| \geq \int_{E_{n,\epsilon}}|f_n - f| = \int |f_n - f| \chi_{E_{n,\epsilon}}\geq \epsilon \mu(E_{n,\epsilon})$$
Hence we have $$\epsilon^{-1}\int |f_n - f| \geq \mu(E_{n,\epsilon})$$ Since $\epsilon$ is arbitrary we have $\epsilon^{-1}\int |f_n - f|\rightarrow 0$ which thus implies that $\mu(E_{n,\epsilon}) = 0$.

I believe your approach is correct. I wrote something up before realizing you had already provided a proof (more-or-less what you have, just by proving the contrapositive):
Let $E_{n,\epsilon}=\{x\colon|f_{n}(x)-f(x)|\geq\epsilon\}$. Suppose
$f_{n}$ does not converge to $f$ in measure so that there exists
an $\epsilon>0$ and $\delta>0$ with $\mu(E_{n,\epsilon})\geq\delta$
for infinitely many $n$. This implies $$\int|f_{n}-f|\geq\int\mathcal{X}_{E_{n,\epsilon}}|f_{n}-f|\geq\epsilon\delta$$
for infinitely many $n$ so that $f_n$ does not converge to $f$ in $L^1$. 
