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Assume $m(E) < \infty$. For two measurable functions $g$ and $h$, define $$\rho(g,h) = \int_E \frac{|g-h|}{1+|g-h|}.$$ Show that $f_n \to f$ in measure on $E$ if and only if $\lim_{n \to \infty} \rho(f_n,f)=0$.

This problem has been asked here); however, I am having trouble understanding Davide's nebulous and somewhat misleading suggestions (at least to my mind), which I now reproduce:

First, a remark: the map $x\mapsto \frac x{1+x}$ is increasing over the set of non-negative real numbers, and is bounded by $1$.

If $f_n\to f$ in measure, fix $\varepsilon$ and integrate over $\{|f_n-f|>\varepsilon\}$ and $\{|f_n-f|\leqslant \varepsilon\}$.

Conversely, if $d(f_n,f)\to 0$, then $\frac{\varepsilon}{1+\varepsilon}\mu\{|f_n-f|>\varepsilon\}\to 0$ (integrating over the set $\{|f_n-f|>\varepsilon\}$).

His first remark is somewhat confusing, for it seems to hint as using the Lebesgue Dominated Convergence theorem for convergence in measure, where $1$ serves as the integrable function that dominates the sequence $\frac{|f_n-f|}{1+|f_n-f|}$ of measurable functions converging to $0$ in measure. Hence,

$$\lim_{n \to \infty} \int_E \frac{|f_n-f|}{1+|f_n-f|} = \int_E 0 = 0$$

Yet he goes on to suggest integrating over $\{|f_n-f| > \epsilon\}$ and $\{|f_n-f| \le \epsilon\}$. Well, I tried this but I couldn't figure out it. For the first integral, we get

$$\int_{\{|f_n-f| > \epsilon\}} \frac{|f_n-f|}{1+|f_n-f|} \le \int_{\{|f_n-f| > \epsilon\}} \le m(\{x : |f_n(x)-f(x)| > \epsilon \} \to 0 \mbox{ as } n \to \infty $$

However, I couldn't figure out how to show that the second integral is zero or at least converges to $0$ as $n \to \infty$. I tried showing that $\int_{\{|f_n-f| \le \epsilon \}} \frac{|f_n-f|}{1+|f_n-f|}$ is bounded above by $\epsilon$ and then letting $\epsilon \to 0^+$, which would actually show that $\int_{\{|f_n-f| \le \epsilon \}} \frac{|f_n-f|}{1+|f_n-f|} = 0$ for every $n$, but I couldn't get just $\epsilon$ alone. The best I could do is get $\epsilon \cdot m(\{x : |f_n(x)-f(x)| \le \epsilon\})$, which doesn't obviously go $0$ as $\epsilon \to 0^+$.

As for the converse, I believe I was able to follow his suggestion. If $|f_n-f| > \epsilon$, then $\frac{|f_n-f|}{1+|f_n-f|} > \frac{\epsilon}{1+\epsilon}$ and therefore $\int_E \frac{|f_n-f|}{1+|f_n-f|} > \int_E \frac{\epsilon}{1 + \epsilon} > \frac{\epsilon}{1+\epsilon} \int_{\{f_n-f| > \epsilon\}} = \frac{\epsilon}{1+\epsilon} m(\{x : |f_n(x)-f(x)| > \epsilon\})$ from which we get

$$0 = \lim_{n \to \infty} \int_E \frac{|f_n-f|}{1+|f_n-f|} \ge \frac{\epsilon}{1+\epsilon} \lim_{n \to \infty} m(\{x : |f_n(x)-f(x)| > \epsilon\})$$

and hence $\lim_{n \to \infty} m(\{x : |f_n(x)-f(x)| > \epsilon\})=0$.

Community
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user193319
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  • $\epsilon\cdot m({x:|f_n(x)−f(x)|≤\epsilon})$ does go to zero as $\epsilon\to 0^+$. This follows since $m({x:|f_n(x)−f(x)|≤\epsilon})\le m(E)$, and the problem assumed $m(E)<\infty$. – Mike Earnest Apr 12 '18 at 15:50
  • @MikeEarnest Oh! Drats! It's so simple! So this means that $\int_{{|f_n-f| \le \epsilon} = 0$ for all $n \in \Bbb{N}$, right? That's somewhat unexpected. – user193319 Apr 12 '18 at 15:58
  • Your last comment is not correct. Why would it be the case that $\int_{{|f_n-f| \le \epsilon}} |f_n-f|/(1+|f_n-f|)= 0$? You are integrating a positive function over a set of positive measure. – Mike Earnest Apr 12 '18 at 16:04

3 Answers3

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Let me give more details. Let $$I(\varepsilon):=\int_{\{|f_n-f| \leqslant\epsilon \}} \frac{|f_n-f|}{1+|f_n-f|}=\int\mathbf 1{\{|f_n-f| \leqslant\epsilon \}} \frac{|f_n-f|}{1+|f_n-f|}.$$ Since the map $x\mapsto x/(1+x)$ is increasing, we certainly have (point wise) $$ \mathbf 1{\{|f_n-f| \leqslant\epsilon \}} \frac{|f_n-f|}{1+|f_n-f|}\leqslant \mathbf 1{\{|f_n-f| \leqslant\epsilon \}} \frac{\epsilon}{1+\epsilon}\leqslant\epsilon \mathbf 1{\{|f_n-f| \leqslant\epsilon \}} $$ hence integrating, we get $$ \int\mathbf 1{\{|f_n-f| \leqslant\epsilon \}} \frac{|f_n-f|}{1+|f_n-f|}\leqslant \varepsilon\cdot m\left(\{|f_n-f| \leqslant\epsilon \}\right)\leqslant \varepsilon \cdot m(E). $$

Davide Giraudo
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Two straightforward inequalities settle the matter. For a measurable function $f$ and $0<\epsilon<1$, $$ m\{x\in E: |f(x)|>\epsilon\}\le{1+\epsilon\over\epsilon}\int_E{|f|\over1+|f|}\,dm $$ and $$ \int_E{|f|\over1+|f|}\,dm\le\epsilon m(E)+m\{x\in E: |f(x)|>\epsilon\}. $$

John Dawkins
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We have by Holder's inequality that \begin{align*} m(\{x\in E: |f_{n}(x)-f(x)|\leq\epsilon\})&=\int_{E}\chi_{|f_{n}-f|\leq\epsilon}\chi_{E}dm\\ &\leq\|\chi_{|f_{n}-f|\leq\epsilon}\|_{L^{\infty}(E)}\|\chi_{E}\|_{L^{1}(E)}\\ &\leq\epsilon\|\chi_{E}\|_{L^{1}(E)}\\ &=m(E)\epsilon, \end{align*} so \begin{align*} \int_{|f_{n}-f|\leq\epsilon}\dfrac{|f_{n}-f|}{1+|f_{n}-f|}dm\leq\int_{|f_{n}-f|\leq\epsilon}dm=m(\{x\in E:|f_{n}-f|\leq\epsilon\})\leq m(E)\epsilon. \end{align*}

user284331
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