# Rudin RCA ch. 1 ex. 7: Monotone convergence theorem for decreasing sequence

Suppose $$f_n:X\rightarrow[0,\infty]$$ is measurable for $$n=1,2,3,\dots$$, $$f_1\geq f_2\geq f_3\geq\cdots\geq 0$$, $$f_n(x)\rightarrow f(x)$$ as $$n\rightarrow\infty$$, for every $$x\in X$$, and $$f_1\in L^1(\mu)$$. Prove that then $$\tag{*}\lim_{n\rightarrow\infty}\int_X f_n\,d\mu=\int_X f\,d\mu$$ and show that this conclusion does not follow if the condition "$$f_1\in L^1(\mu)$$" is omitted.

Let $$E$$ consist of the points $$x\in X$$ at which $$f_1(x)<\infty$$. By the dominated convergence theorem, $$\int_E f_n\,d\mu\rightarrow \int_E f\,d\mu\mbox{.}$$ Since $$f_1\in L^1(\mu)$$, $$\mu(E^c)=0$$, and hence (*) follows.

Let $$X=\{1,2,3,\dots\}$$, and let $$\mu$$ be the counting measure. For each $$n$$, define $$f_n:X\rightarrow[0,\infty]$$ by $$f_n(x)=\left\{\begin{array}{ll}\infty&(x\geq n)\\0&(x Then $$\lim f_n=0$$, and $$\int_X f_n\,d\mu=\infty$$ for all $$n$$.

Is this correct?

• I don't think you can use the dominated convergence theorem right away, as this requires a monotonically increasing sequence of functions. You can save your proof by looking at functions $g_k := f_1 - f_k$. Commented Mar 2, 2022 at 8:04
• Does this answer your question? monotone convergence theorem for decreasing sequence. Commented Mar 2, 2022 at 15:57

Since $$0\leq f\leq f_n\leq f_1 \in L^1,\,\forall n$$ and $$f_n\downarrow f$$ then you can use DCT and the conclusion follows. However that's probably not what you're supposed to do. As suggested in the comments, the sequence given by $$u_k:=f_1-f_k$$ is nonnegative and increasing s.t. you can use MCT for nonnegative increasing sequences $$\sup_n\int (f_1-f_n)d\mu=\int (f_1-f)d\mu$$ Since $$f_n,f_1,f\in L^1$$ we get $$\sup_n\int (-f_n)d\mu=\sup_n\int((f_1-f_n)-f_1)d\mu=\int (f_1-f)d\mu-\int f_1d\mu=\int (-f)d\mu$$ and the claim follows since $$\sup_n(-\int f_n)=-\inf_n\int f_n$$.
• How do you avoid $\infty-\infty$?
• The second equality of the second display doesn't make any sense to me. For $f_1-f_n$ may be $\infty$ and Rudin's text only proves that $\int f-g=\int f-\int g$ if $f, g$ are complex-valued.
• @user912011 if $N$ is a null set, then $\int_N u d\mu=0$ Commented Mar 3, 2022 at 1:54