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.
$f_n(x)$ convergence in measure implies $\frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}$ convergence almost everywhere
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0Almost everywhere convergence does not imply convergence in measure. (Take $f_n = 1_{[n, \infty)}$.) – 2011-12-30
1 Answers
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.
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5It 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). – 2011-12-31