Let \(f_1(x)\le f_2(x)\le\cdots\le f_n(x)\le\cdots\) on a set \(A\), where the functions \(f_n\) are integrable and \(\int_A f_n(x)\,dx\le M\) for some constant \(M\). Show that the limit \[ f(x)=\lim_{n\to\infty}f_n(x) \] exists and is finite almost everywhere on \(A\), and that \[ \lim_{n\to\infty}\int_A f_n(x)\,dx=\int_A f(x)\,dx. \]
Solution
Set \[ g_n:=f_n-f_1. \] Then each \(g_n\) is measurable, nonnegative, and \[ 0=g_1\le g_2\le\cdots. \] Moreover \[ \int_A g_n =\int_A f_n-\int_A f_1 \le M-\int_A f_1<\infty. \]
Let \[ g(x):=\lim_{n\to\infty}g_n(x)\in[0,\infty]. \] By the Monotone Convergence Theorem, \[ \int_A g =\lim_{n\to\infty}\int_A g_n \le M-\int_A f_1<\infty. \] Hence \(g(x)<\infty\) for almost every \(x\in A\). Therefore \[ f(x):=f_1(x)+g(x) =\lim_{n\to\infty}f_n(x) \] exists and is finite almost everywhere.
Finally, \[ \begin{aligned} \lim_{n\to\infty}\int_A f_n &=\int_A f_1+\lim_{n\to\infty}\int_A g_n\\ &=\int_A f_1+\int_A g\\ &=\int_A f. \end{aligned} \] Thus the pointwise monotone limit is finite almost everywhere and the integrals converge to the integral of the limit.