$(f_n)$ a decreasing sequence of integrable function, prove $\int lim f_n \ d\mu = lim \int f \ d\mu$

measure-theory

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?

Best Answer

$-f_n + f_1$ 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.

Best Answer

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