Let $(X, A,μ)$ be a complete measure space and let $(f_n(x))_{n=1}^{\infty}$ be a sequence of nonnegative measurable functions defined on a measurable set $E$ such that:
1) $0≤f_1(x)≤f_2(x)≤…≤f_n(x)≤…≤f(x)$ on $E$.
2) $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$.
Then $\lim_\limits{n→∞}∫_{E}f_n(x)dμ=∫_{E}f(x)dμ$.
Proof from a website.) Since for each $n\in\mathbb{N}$, $0\leq f_n(x)≤f(x), ∫_{E}f_n(x)dμ\leq∫_{E}f(x)dμ$ and
$\lim_\limits{n→∞}∫_{E}f_n(x)dμ\leq∫_{E}f(x)dμ$. We also note that $(∫_{E}f_n(x)dμ)_{n=1}^{\infty}$ is an increasing sequence of nonnegative real numbers that is bounded above, and hence converges, so $\lim_\limits{n→∞}∫_{E}f_n(x)dμ$ converges.Question.) Can we always assume $∫_{E}f(x)dμ<\infty$? I think if
$∫_{E}f(x)dμ=\infty$, $(∫_{E}f_n(x)dμ)_{n=1}^{\infty}$ may not be a
bounded sequence.
Best Answer
I suppose this is not really an answer to the specific question about what appears on that website; including a link to said site might have been a good idea. (Speaking of good ideas, using Fatou's lemma here may not be one, since FL is typically proved as a corollary to MCT...)
It is an answer to the related question "How do I prove MCT?". The standard proof is very simple, and works regardless of whether $\int f$ is finite. Simple enough that I should be able to reproduce it, having the idea in my head; let's see:
We may as well assume $E=X$. Suppose $\phi$ is a simple function with $0\le\phi\le f$. Say $$\phi=\sum_{j=1}^Na_j\chi_{E_j},$$where $a_j>0$ and the $E_j$ are disjoint.
Fix $\lambda\in(0,1)$. Let $$E_{j,n}=\{x\in E_j:f_n(x)>\lambda\phi(x)\}.$$Since $\lambda\phi(x)<f(x)$ for $x\in E_j$ we have $$E_j=\bigcup_nE_{j,n},$$so, since $E_{j,n}\subset E_{j,n+1}$ we have $$\lim_n\mu(E_{j,n})=\mu(E_j).$$But $\int f_n\ge\sum_j\lambda a_j\mu(E_{j,n})$, so $$\lim_n\int f_n\ge\lim_n\sum_j\lambda a_j\mu(E_j)=\lambda\int\phi.$$Hence $$\lim_n\int f_n\ge\sup_{\lambda\in(0,1)}\lambda\int\phi=\int\phi,$$so the definition of $\int f$ shows that $$\lim_n\int f_n\ge\sup_\phi\int\phi=\int f.$$