I have not posted in a gargantuan amount of time, so here is a quick result:

Claim: Let $(X,\mathcal M,\mu)$ be a measure space, let $f_n,f\in L^1(\mu)$ for all $n\in\mathbb N$, and let $f_n\to f$ in $L^1(\mu)$. Then $f_n\to f$ in measure.

Proof: Let $E_{\varepsilon,n}=\{x\in X:|f_n(x)-f(x)|\geq\varepsilon\}$, where $\varepsilon>0$ and $n\in\mathbb N$. Then $$\int|f_n-f|d\mu\geq\int_{E_{\varepsilon,n}}|f_n-f|d\mu\geq\varepsilon\mu(E_{\varepsilon,n}).$$ Therefore, $$\mu(E_{\varepsilon,n})\leq\varepsilon^{-1}\int|f_n-f|\xrightarrow{n\to\infty}0.\tag*{\blacksquare}$$
The converse of this claim does not hold:

Let $f_n=n\chi_{[0,n^{-1}]}$, let $\varepsilon>0$, and let $n\in\mathbb N$. Then
$$\mu(\{x\in X:n\chi_{[0,n^{-1}]}(x)\geq\varepsilon\})\xrightarrow{n\to\infty}0$$
since $\mu(\{x\in X:n\chi_{[0,n^{-1}]}(x)\geq\varepsilon\})\leq n^{-1}$. Therefore, $f_n$ converges to $0$ in measure. However,
$$\int n\chi_{[0,n^{-1}]}d\mu=n\mu([0,n^{-1}])=1.$$