Prove the monotone convergence theorem for sequences of Lebesgue-integrable functions

Question

I'm trying to prove the monotone convergence theorem for $L^1$ functions: Suppose $(f_n)$ is a sequence of $L^1$-functions (i.e Lebesgue-integrable functions) over a measure space $(X,\sigma (X), \mu)$ such that $f_n(x)\le f_{n+1}(x)$, $\forall n=1,2,...$, for $x\in X$ almost everywhere (a.e.). If $$\lim_{n \to \infty} \int_X f_n \,\text{d}\mu=c\in \Bbb R$$, show that there is a function $f\in L^1(X)$ such that $$\lim_{n \to \infty}f_n(x)=f(x) \,\text{a.e.}$$ and $$\int_X f\,\text{d}\mu=c$$ I tried to prove this by Lebesgue dominated convergence theorem, but got stuck proving the above statement. Could someone help to provide a proof please? Thanks a lot.

Answer 1

The sequence $g_n:=f_n-f_1\ge 0$ is non-decreasing, so $g(x):=\lim_ng_n(x)$ exists for all $x$, and (by Fatou's lemma) $\int_X g\,d\mu\le\liminf_n\int_X g_n\,d\mu =c-\int_X f_1\,d\mu<\infty$. Therefore $g\ge 0$ is integrable. Of course, $f_n$ increases pointwise to $f=g+f_1$ (which is also integrable). Finally, $f_1\le f_n\le f$, so $|f_n|\le |f_1|+|f|$, and by Dominated Convergence, $\int_X f\,d\mu=\lim_n\int_X f_n\,d\mu = c$. (There is no need to assume $f_n\ge 0$.)

Answer 2

$\{f_n\}$ is monotone, then $f_n(x)$ converges to some number (denote as $f(x)$) for a.e. $x$

One side of inequality comes from Fatou's lemma. $f_n \geq 0$, $f_n \rightarrow f$, apply Fatou's lemma $$\liminf_n \int f_n \geq \int f.$$

The other side comes from monotoncity $f_n \leq f$, then $\int f_n \leq \int f$, and $$\limsup_n \int f_n \leq \int f.$$