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

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.

• This is false unless you assume $f_nn\ge0$. Given that assumption it's trivial from DCT; what part are you stuck on? Apr 29, 2016 at 14:37
• Yes, I forgot to assume $f_n \ge 0$. Since I also noticed a theorem that if we let ${g_n}$ be a sequence of $L(I)$ functions such that $g_n\ge 0$ a.e. and the series $\sum_{n=1}^\infty \int_Ig_n$ converges, then the series $\sum_{n=1}^\infty g_n$ converges a.e. on $I$ to a sum function $g$ in $L(I)$, and have $\int_Ig=\int_I\sum_{n=1}^\infty g_n=\sum_{n=1}^\infty\int_Ig_n$. I'm not sure if I have to prove this first and then derive the result, or is there a way to prove the statement by applying DCT directly. Apr 29, 2016 at 14:54
• It follows directly from DCT. (I don't see how that other result you mention could possibly apply, since the partial sums of the series are increasing, and your sequence is decreasing.) Apr 29, 2016 at 14:56

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 $f_n\ge 0$.)
• When you apply Fatou's lemma, are you assuming $g$ is measurable? Apr 30, 2016 at 17:22
• As $g$ is the pointwise limit of a sequence of measurable functions, it is necessarily measurable. Apr 30, 2016 at 17:25
$\{f_n\}$ 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 $$\limsup_n \int f_n \leq \int f.$$