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Dominated Convergence Theorem: Let $f_n$ be measurable and assume that $f_n$ converges a.e. to $f$. If $|f_n(x)|\leq g(x)$ for some integrable $g$ it follows that $f_n$ converges in $L^1$ to $f$, in particular $\int f_n\to\int f$. Hence, we do not only have the convergence of the integrals but even $L^1$-convergence.

Now my question: Does the monotone convergence theorem also implies $L^1$ convergence or just $\int f_n\to\int f$?

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  • $\begingroup$ Hint $$\int_{\mathbb R}|f_n-f|=\int_{\mathbb R}f-\int_{\mathbb R}f_n.$$ $\endgroup$ – Surb Sep 29 '18 at 10:26
  • $\begingroup$ So the answer must be yes since the right side converges to 0 by MCT. In the literature it is often noticed that DCT provide $L^1$-convergence, but never for MCT. So I thought that this will not hold in general. $\endgroup$ – RandomUser Sep 29 '18 at 10:32
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    $\begingroup$ The problem is that $\int f$ can by infinite. So, indeed MCT do not provide $L^1$ convergence. But if $\int f<\infty$, then $f_n\to f$ in $L^1$. $\endgroup$ – Surb Sep 29 '18 at 11:27
  • $\begingroup$ Thank you, that makes sense. $\endgroup$ – RandomUser Sep 29 '18 at 11:45

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