Let $(\Omega,\mathcal{F},\mu)$ be a measure space. I know the following result:
Let $\{X_n\}_{n=1}^{\infty}$ be an increasing sequence of nonnegative measurable functions with $X=\lim_nX_n$. Then $\int_{\Omega}X\,d\mu=\lim_n\int_{\Omega}X_n\,d\mu$.
I want to prove: Let $\{X_n\}_{n=1}^{\infty}$ an increasing sequence of integrable functions with $X=\lim_nX_n$. If $\sup_n\int_{\Omega}X_n\,d\mu<\infty$, then $X$ is integrable and $\int_{\Omega}X\,d\mu=\lim_n\int_{\Omega}X_n\,d\mu$.
My attempt: We have that $\{X_n-X_1\}_{n=1}^{\infty}$ is an increasing sequence of nonnegative measurable functions. By the previous result, $$\int_{\Omega}(X-X_1)\,d\mu=\lim_n\int_{\Omega}(X_n-X_1)\,d\mu=\lim_n\int_{\Omega}X_n\,d\mu-\int_{\Omega}X_1\,d\mu.$$ If I assume the integrability of $X$, I can say $$\int_{\Omega}(X-X_1)\,d\mu=\int_{\Omega}X\,d\mu-\int_{\Omega}X_1\,d\mu,$$ so $$\int_{\Omega}X\,d\mu=\lim_n\int_{\Omega}X_n\,d\mu.$$ Thus, it remains to prove that $X$ is integrable. Here is where I should use the fact that $\sup_n\int_{\Omega}X_n\,d\mu<\infty$, but I do not know how to proceed.
I will give my own answer. From $$\int_{\Omega} \underbrace{(X-X_1)}_{\geq0}\,d\mu=\lim_n\int_{\Omega}X_n\,d\mu-\int_{\Omega}X_1\,d\mu<\infty,$$ we deduce that $X-X_1$ is integrable. Then $X=(X-X_1)+X_1$ is integrable.