# $f_n(x)$ convergence in measure implies $\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}$ convergence almost everywhere

Let $E\subset\mathbb{R},m(E)<+\infty$, $\{f_n(x)\}$ are measurable functions defined on $E$. Then $\{f_n(x)\}$ converges to $f(x)$ in measure $\Leftrightarrow$ $$\lim_{n\rightarrow\infty}\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}=0, a.e. x\in E.$$ I think it's easy to see how to go from right to left since $\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}<\epsilon\Rightarrow|f_n(x)-f(x)|<2\epsilon$ and almost everywhere convergence implies convergence in measure. What baffles me is the other direction.

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Have you copied the problem correctly? The result you have written is false. – Byron Schmuland Dec 29 '11 at 20:43
I'm sure I copied it correctly. There is even a short hint which I'm not able to follow. I would be more than surprised if it's true. – Tim Dec 29 '11 at 20:59
See the standard counterexample for convergence in measure not implying almost everywhere convergence, e.g. at en.wikipedia.org/wiki/Convergence_in_measure – Robert Israel Dec 29 '11 at 21:01
Almost everywhere convergence does not imply convergence in measure. (Take $f_n = 1_{[n, \infty)}$.) – Nate Eldredge Dec 30 '11 at 1:05

I guess the result you have to show is that the convergence in measure of $\{f_n\}$ to $f$ is equivalent to $$\lim_{n\to\infty}\int_E\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}dm (x).$$ (your result is not true, because $\displaystyle\lim_{n\to\infty}\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}$ is equivalent to $\lim\limits_{n\to\infty}|f_n(x)-f(x)|=0$ and there are sequences which converge in measure but not almost everywhere.)
To show the result, if $f_n\to f$ in measure, then fix $\varepsilon>0$. Since $t\mapsto \frac t{t+1}$ is increasing $$\int_E\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}dm (x)\leq \frac{\varepsilon}{1+\varepsilon}m(E)+m(|f_n-f|\geq \varepsilon),$$ so $$\limsup_{n\to\infty}\int_E\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}dm (x)\leq \frac{\varepsilon}{1+\varepsilon}m(E),$$ and $$\lim_{n\to\infty}\int_E\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}dm (x)=0.$$ Conversely, $$\frac{\varepsilon}{1+\varepsilon} \cdot m(|f_n-f|\geq\varepsilon)\leq \int_{\left\{|f_n-f|\geq\varepsilon\right\}}\frac{\varepsilon}{1+\varepsilon}dm(x)\leq \int_{\left\{|f_n-f|\geq\varepsilon\right\}}\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}dm(x),$$ and we are done.
It might be worth pointing out that $$d(f,g) = \int_{E} \frac{|f-g|}{1+|f-g|}\,dm$$ is a metric on the space of measurable functions $f: E \to \mathbb{R}$ modulo null-functions and that this space is complete in this metric. Incidentally, it is one of the very few non-locally convex topological vector spaces that are of actual use (in fact, the only convex neighborhood of $0$ is the space itself). – t.b. Dec 31 '11 at 0:01