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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real-analysis
measure-theory
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2Have you copied the problem correctly? The result you have written is false. – 2011-12-29
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0I'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. – 2011-12-29
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1See the standard counterexample for convergence in measure not implying almost everywhere convergence, e.g. at http://en.wikipedia.org/wiki/Convergence_in_measure – 2011-12-29
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0Almost everywhere convergence does not imply convergence in measure. (Take $f_n = 1_{[n, \infty)}$.) – 2011-12-30