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

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?

$$-f_n + f_1$$ 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.
• $f_1 \in L \iff \int f_1 \ d\mu < \infty$... but $\int f_n \ d\mu \leq \int f_1 \ d\mu <\infty$.. whats wrong? – Robson Oct 7 '18 at 2:13
• e.g. $f_n=-\infty$ – Hasse1987 Oct 7 '18 at 2:23
• @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. – Hasse1987 Oct 7 '18 at 3:55