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I am reading Measure Theory by Donald L. Cohn and practicing my understanding on exercices. I find this one challenging :

Let ($X,\mathcal{A},\mu$) be a measure space and let $f : X → \bar{\mathbb{R}}$ be an $\mathcal{A}$-measurable function. Suppose that nested sets $A_1$$A_2$ ⊆ ..., all in $\mathcal{A}$, satisfy $\cup_{n=1}^{\infty}A_n=X$ and $lim_{n→∞}\int_{A_{n}}|f|d\mu<∞$. Show that f is $\mu$-integrable.

Thank you for your help !

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$\int_{A_n}|f|d\mu=\int_X 1_{A_n}|f|d\mu$.

Since $1_{A_n}|f|$ increases and pointwisely converges to $1_{\cup_{n=1}^{\infty}A_n}|f|=1_X |f|=|f|$,

$\int_X 1_{A_n}|f|d\mu$ increases and converges to $\int_X |f|d\mu$, by a standard theorem of measure theory. By the hypothesis this limit is finite, i.e. $f$ is $\mu$-integrable.

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  • $\begingroup$ Hello, thank you for your answer ! This is what I thought about doing but couldn't find the theorem needed in Cohn's book. Could you please give me the name of the theorem ? $\endgroup$ – 15462764 Nov 19 '19 at 16:00
  • $\begingroup$ It is frequently called (Beppo Levi's) Monotone Convergence Theorem $\endgroup$ – Sea watcher Nov 20 '19 at 2:52

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