$\rho(f,g)=\int_E \min(1,|f-g|)dm$. Prove that $f_n$ converges to $f$ in measure if and only if $\rho(f_n,f)\rightarrow 0$ as $n\rightarrow\infty$

Question

Question: Suppose $m$ is a finitemeasure on a measurable space $E$. Define $\rho(f,g)=\int_E \min(1,|f-g|)dm$. Prove that $f_n$ converges to $f$ in measure if and only if $\rho(f_n,f)\rightarrow 0$ as $n\rightarrow\infty$.

My Thoughts: For the backwards direction, if $\rho(f_n,f)\rightarrow 0$ as $n\rightarrow 0$, then we have that $\lim_{n\rightarrow\infty}\int_E\min(1,|f_n-f|)dm=\lim_{n\rightarrow\infty}\int_E|f_n-f|dm=0$, and so we have uniform convergence, which implies in measure convergence. For the forward direction, I am not really sure if I should consider cases, that is, when $1$ is the minimum and then when $f_n-f$ is the minimum and show that if $1$ is the minimum then the statement can't be true, and so $f_n-f$ must be the minimum and then show that the only way we get in measure convergence is if that integral equals $0$... but I am not quite sure how to do that. Any thoughts, suggestions, etc. are greatly appreciated! Thank you.

Answer 1

In the backwards direction you don't have uniform convergence, only that $\int_E \min(1,|f_n-f|)\,dm \to 0$. Also, the conclusion $$\int_E \min(1,|f_n-f|)\,dm \to 0 \implies \int_E |f_n-f|\,dm \to 0$$ is not entirely spelled out in your question.

Anyway, for $0 < \varepsilon < 1$ you can use Markov's inequality to obtain $$m(|f_n-f| > \varepsilon) =m\Big(\min(1,|f_n-f|) > \varepsilon\Big)\le \frac1\varepsilon \int_E\min(1,|f_n-f|)\,dm \xrightarrow{n\to\infty} 0.$$

Conversely, assume that $f_n \to f$ in measure $m$ and let $0 < \varepsilon <1$. Pick $n_0 \in \Bbb{N}$ such that for $n \ge n_0$ we have $$m\left(|f_n-f| >\frac\varepsilon{2m(E)}\right) < \frac\varepsilon2.$$ Then for all $n \ge n_0$ we have \begin{align} \int_E \min(1,|f_n-f|)\,dm &\le \int_{|f_n-f| \le \frac\varepsilon{2m(E)}} |f_n-f|\,dm + \int_{|f_n-f| > \frac\varepsilon{2m(E)}} 1\,dm\\ &\le \frac\varepsilon{2m(E)}m\left(|f_n-f| \le\frac\varepsilon{2m(E)}\right) + m\left(|f_n-f| >\frac\varepsilon{2m(E)}\right)\\ &\le \frac\varepsilon2 + \frac\varepsilon2\\ &=\varepsilon \end{align} so we conclude $\int_E \min(1,|f_n-f|)\,dm \to 0$.

Answer 2

$\forall\epsilon>0,\exists N$ s.t. $\forall n\ge N,m(\{x\in E\mid|f_n(x)-f(x)|>\epsilon\})<\epsilon$

$\Rightarrow \forall n\ge N,\rho(f_n,f)=\int_E \min\{1,|f_n-f|\}dm$ $=\int_{\{x\in E\mid|f_n(x)-f(x)|>\epsilon\}}1dm+\int_{E\setminus\{x\in E\mid|f_n(x)-f(x)|>\epsilon\}}|f_n-f|dm<\epsilon+\epsilon\cdot m(E)$

Thus, convergence in measure implies convergence wrt $\rho$.

Conversely,

$\forall\epsilon>0,\exists N$ s.t. $\forall n\ge N,\rho(f_n,f)=\int_E\min\{1,|f_n-f|\}dm<\epsilon$

$\Rightarrow m(\{x\in E\mid|f_n(x)-f(x)|>\epsilon\})\le\int_{\{x\in E\mid|f_n(x)-f(x)|>\epsilon\}}1dm\le\int_E\min\{1,|f_n-f|\}dm<\epsilon$

Thus, convergence wrt $\rho$ implies convergence in measure as well.

Remark. While the definition of convergence measure is stated in a slightly different way (undistinguishing $\epsilon$), you can easily notice that they are the same ones.