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Let $(\Omega ,F,\mu)$ a be a measure space & $\{f_n\}$ a sequence of nonnegative Lebesgue integrable functions. If $\{f_n\}$ converges in measure to function $f $ & the integrals converges to $\int_\Omega f < \infty $, prove that $\int_\Omega |f_n -f|\,d\mu$ converges to $0$, when $n $ goes to infinity.

It seems easy, but I didn't success.

Help me please.

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Since $\{f_n\}$ is a sequence of nonnegative Lebesgue integrable functions and $\{f_n\}$ converges in measure to function $f $, we have that $\mu([f<0])=0$. So $f\geqslant 0$ a.e.. So, without loss of generality, we can assume that $f\geqslant 0$.

Now consider $f_n \wedge f$ defined by $(f_n \wedge f)(x)=\min\{f_n(x),f(x)\}$, for each $x \in \Omega$.

Since for any $\varepsilon>0$, $$ [\vert (f_n \wedge f)-f \vert\geqslant \varepsilon ]= [f-(f_n \wedge f)\geqslant \varepsilon ] = [f-f_n \geqslant \varepsilon ] \subseteq [\vert f_n -f \vert \geqslant \varepsilon ] $$ we have that $\{f_n \wedge f\}$ converges in measure to function $f$. But we know that, for all $n$, $\vert f_n \wedge f \vert = f_n \wedge f \leqslant f $ and $\int_{\Omega } f d\mu< \infty $. So we can apply the Dominated Convergence Theorem (for convergence in measure) and we have that $$\lim_{n \to \infty}\int_{\Omega } f_n \wedge f d\mu = \int_{\Omega } f d\mu$$ To conclute the proof, note that $$\vert f_n-f\vert = f_n+f-2(f_n\wedge f)$$ So $$\int_{\Omega } \vert f_n-f\vert d\mu = \int_{\Omega } f_n d\mu +\int_{\Omega } f d\mu -2\int_{\Omega } (f_n\wedge f) d\mu $$ And so, since $\lim_{n \to \infty}\int_{\Omega } f_n d\mu = \int_{\Omega } f d\mu$ we have $$ \lim_{n \to \infty}\int_{\Omega } \vert f_n-f\vert d\mu =0$$

Remark: Note that it was NOT assumed that $\mu(\Omega)<+\infty$.

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  • $\begingroup$ Yeah, this is correct. You don't even need $\sigma$-finite! $\endgroup$ – user24142 Oct 26 '15 at 6:43
  • $\begingroup$ @hermes There are two versions of the Dominated Convergence Theorem. One version for $f_n \to f$ pointwise a.e. and another version for $f_n \to f$ in measure. That is why I explictly stated that I was applying the Dominated Convergence Theorem for convergence in measure. (See Halmos $\S$26 Theorem D or Dunford-Schwartz III.3.7). $\endgroup$ – Ramiro Oct 29 '15 at 0:03
  • $\begingroup$ It needs $\sigma$-finite too for Dominated Convergence Theorem for convergence in measure. $\endgroup$ – hermes Oct 29 '15 at 0:12
  • $\begingroup$ @hermes No, it does NOT require $\sigma$-finite for Dominated Convergence Theorem for convergence in measure. Please, see the references: Halmos $\S$26 Theorem D and Dunford-Schwartz III.3.7. $\endgroup$ – Ramiro Oct 29 '15 at 0:19

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