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 $\mu(E_{n,\epsilon}) = 0$.

@Wolfy ,  the proof is actually simpler than you wrote. And actually it needs some adjustments. I have copied your proof, making the necessary adjustments.

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

Proof - 
Given any $\epsilon>0$. Let $E_{n,\epsilon} = \{x:|f_n(x) - f(x)|\geq \epsilon\}$. Since $|f_n - f|\geq 0$, we have, for all $n$, 
$$\int |f_n - f| \geq \int_{E_{n,\epsilon}}|f_n - f| = \int |f_n - f| \chi_{E_{n,\epsilon}}\geq \epsilon \mu(E_{n,\epsilon})\geq 0$$
Hence we have $$\epsilon^{-1}\int |f_n - f| \geq \mu(E_{n,\epsilon})\geq 0$$ 
Since  $\int |f_n - f|\rightarrow 0$, we have that $\epsilon^{-1}\int |f_n - f|\rightarrow 0$. So $ \mu(E_{n,\epsilon}) \to 0$.
It means that $ \mu(\{x:|f_n(x) - f(x)|\geq \epsilon\}) \to 0$, as $n \to \infty$, which means (by definition) that $f_n \to f$ in measure.
