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.
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$\begingroup$ This is false unless you assume $f_nn\ge0$. Given that assumption it's trivial from DCT; what part are you stuck on? $\endgroup$– David C. UllrichCommented Apr 29, 2016 at 14:37
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$\begingroup$ 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. $\endgroup$– user3029Commented Apr 29, 2016 at 14:54
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$\begingroup$ 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.) $\endgroup$– David C. UllrichCommented Apr 29, 2016 at 14:56
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2 Answers
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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 $f_n\ge 0$.)
answered Apr 30, 2016 at 0:51
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$\begingroup$ When you apply Fatou's lemma, are you assuming $g$ is measurable? $\endgroup$– user3029Commented Apr 30, 2016 at 17:22
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1$\begingroup$ As $g$ is the pointwise limit of a sequence of measurable functions, it is necessarily measurable. $\endgroup$ Commented Apr 30, 2016 at 17:25
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$\{f_n\}$ 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 $$\limsup_n \int f_n \leq \int f.$$
