If $(X,\tau,\mu)$ measure space and $(f_n)_n$ measurable functions, and $f$ measurable function. If $f_n\to f$ a.e. then $f\to f$ in measure? or we need $\mu$ is finite?
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I want to say yes? – Sean Roberson Jul 10 '18 at 05:58
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For a counter example when $\mu $ is not finite take $f_n=I_{(n,\infty )}$ and $\mu$ equal to Lebesgue measure. – Kavi Rama Murthy Jul 10 '18 at 07:45
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By Fatou's lemma :
$$\lim \inf \int_X\textbf{1}_{\{x, |f_n(x)-f(x)|< \varepsilon\}} d \mu \ge \int_X\lim \inf\textbf{1}_{\{x, |f_n(x)-f(x)|< \varepsilon\}} d \mu$$
If $X$ has finite measure, then, by a.e. convergence : $$\lim \inf \int_X\textbf{1}_{\{x, |f_n(x)-f(x)|< \varepsilon\}} d \mu \ge \mu(X).$$
Thus equality. Then: $\int_X\textbf{1}_{\{x, |f_n(x)-f(x)| \ge \varepsilon\}} d \mu=0$ and $f_n$ converges in measure to $f$.
If $X$ does not have finite measure, we cannot conclude without further hypothesis. Consider $f_n(x)=\textbf{1}_{(n-\frac 1 2,n+\frac 1 2)}$ on $X=\mathbb{R}$ with the Lebesgue measure. This function converges a.e. to the null function but we have :
$$\forall n, \int_X\textbf{1}_{\{x, |f_n(x)-f(x)|\ge \varepsilon\}} d \mu=1.$$
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