# 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?

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, $$\lim_\limits{n\to\infty} \inf\limits_{\alpha>0} \{\alpha+m(\{x \in E: |f_n(x)-f(x)|>\alpha\})\}\neq 0$$