Equivalent condition of convergence in measure Suppose $E \subset \Bbb R^n$, Lebesgue mesure $m(E) < +\infty$.
$f(x), f_1(x), \cdots, f_n(x), \cdots$ are measurable functions on $E$.
Prove that

$f_n$ convergence to $f$ in measure $\iff$ $\lim_\limits{n\to\infty} \inf\limits_{\alpha>0} \{\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})\}=0$.

($\implies$) $\forall \varepsilon > 0$, $\alpha<\frac{\varepsilon}{2}$, there exists $N$ s.t. for $n \ge N$, $m(\{x \in E: |f_n(x)-f(x)|>\alpha\})<\frac{\varepsilon}{2}$, so $\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})<\varepsilon$ and we can take the infimum.
However, for the ($\impliedby$) part, $\alpha$ appears in  form $\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})$, so it's difficult to single out the $m(\{x \in E: |f_n(x)-f(x)|>\delta\})$ term. So how can we proceed?
 A: For the ($\implies$) part, your argument is correct. Let us now prove:

$f_n$ convergence to $f$ in measure $\iff$ $\lim_\limits{n\to\infty} \inf\limits_{\alpha>0} \{\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})\}=0$.

Proof:
($\implies$) As your wrote:  $\forall \varepsilon > 0$, $\alpha<\frac{\varepsilon}{2}$, there exists $N$ s.t. for $n \ge N$, $m(\{x \in E: |f_n(x)-f(x)|>\alpha\})<\frac{\varepsilon}{2}$, so $\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})<\varepsilon$ and we can take the infimum.
($\impliedby$) Let us prove by the counter-positive. Suppose  $f_n$ does not convergence to $f$ in measure. Then, there is $\alpha_0 >0$ and  there is $\varepsilon_0>0$, such that, for all $N\in \Bbb N$, there is $n>N$ such that
$$m(\{x \in E: |f_n(x)-f(x)|>\alpha_0\})>\varepsilon_0$$
So, for all $N\in \Bbb N$, there is $n>N$ such that, for all $\alpha>0$,
if $\alpha > \alpha_0$, then
$$\alpha + m(\{x \in E: |f_n(x)-f(x)|>\alpha\}) \geqslant \alpha > \alpha_0 \geqslant \min(\alpha_0, \varepsilon_0)$$
if $\alpha \leqslant \alpha_0$, then $m(\{x \in E: |f_n(x)-f(x)|>\alpha\}) \geqslant m(\{x \in E: |f_n(x)-f(x)|>\alpha_0\})$. So
\begin{align*} 
\alpha +  m(\{x \in E: |f_n(x)-f(x)|>\alpha\} & \geqslant m(\{x \in E: |f_n(x)-f(x)|>\alpha\}) \geqslant \\
& \geqslant m(\{x \in E: |f_n(x)-f(x)|>\alpha_0\})> \\
& > \varepsilon_0 \geqslant \min(\alpha_0,\varepsilon_0)
\end{align*}
So, for all $N\in \Bbb N$, there is $n>N$ such that, for all $\alpha>0$,
$$\alpha + m(\{x \in E: |f_n(x)-f(x)|>\alpha\})> \min(\alpha_0, \varepsilon_0)$$
So, for all $N\in \Bbb N$, there is $n>N$ such that,
$$\inf_{\alpha>0} \{\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})\} \geqslant \min(\alpha_0, \varepsilon_0)$$
So,
$$\lim_\limits{n\to\infty} \inf\limits_{\alpha>0} \{\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})\}\neq 0$$
