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Let $f_n: X\to \mathbb{R}$ be measurable for all natural $n$ such that $f_1\geq f_2 \geq \cdots$ and $\lim f_n = f$. Prove that if $f_1$ is integrable than $$\int f \ d\mu = \lim \int f_n \ d\mu$$

I know that $f_1$ integrable $\implies f_n$ integrable $\forall n$

Then I tried to use LDCT but I couldn't find a integrable function wich dominates $\vert f_n \vert$... I tried $\max\{\vert f_1\vert,\vert f \vert \}$ but I don't know if $f$ is integrable ($\max$ of integrable $f,g$ is integrable because $\max\{f,g\}\leq\vert f \vert + \vert g \vert$, right?)...

any help?

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$-f_n + f_1$ is increasing and non-negative. You can apply monotone convergence to that sequence then work backwards using the linearity of the integral and integrability of $f_1$. BTW I don't think your assertion about all $f_n$ being integral is accurate.

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  • $\begingroup$ $f_1 \in L \iff \int f_1 \ d\mu < \infty $... but $\int f_n \ d\mu \leq \int f_1 \ d\mu <\infty$.. whats wrong? $\endgroup$ – Robson Oct 7 '18 at 2:13
  • $\begingroup$ e.g. $f_n=-\infty$ $\endgroup$ – Hasse1987 Oct 7 '18 at 2:23
  • $\begingroup$ they are real functions $\endgroup$ – Robson Oct 7 '18 at 2:29
  • $\begingroup$ @Robson Then $f_n=-1/n$ on the unit interval. My point is you can only get that the positive part has finite integral from these assumptions, not the negative part. You need both to conclude integrability. $\endgroup$ – Hasse1987 Oct 7 '18 at 3:55

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