3
$\begingroup$

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 $$f_n(x)=\left\{\begin{array}{ll}\infty&(x\geq n)\\0&(x<n).\end{array}\right.$$ Then $\lim f_n=0$, and $\int_X f_n\,d\mu=\infty$ for all $n$.

Is this correct?

$\endgroup$
2
  • 1
    $\begingroup$ 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 $g_k := f_1 - f_k$. $\endgroup$
    – jasnee
    Commented Mar 2, 2022 at 8:04
  • 3
    $\begingroup$ Does this answer your question? monotone convergence theorem for decreasing sequence. $\endgroup$
    – Mittens
    Commented Mar 2, 2022 at 15:57

1 Answer 1

1
$\begingroup$

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 $\sup_n(-\int f_n)=-\inf_n\int f_n$.

$\endgroup$
4
  • $\begingroup$ How do you avoid $\infty-\infty$? $\endgroup$
    – user912011
    Commented Mar 3, 2022 at 0:58
  • $\begingroup$ wherever that happens it is in a null set (because all functions involved are integrable), and there the integral is zero @user912011 $\endgroup$
    – Snoop
    Commented Mar 3, 2022 at 1:36
  • $\begingroup$ 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. $\endgroup$
    – user912011
    Commented Mar 3, 2022 at 1:45
  • $\begingroup$ @user912011 if $N$ is a null set, then $\int_N u d\mu=0$ $\endgroup$
    – Snoop
    Commented Mar 3, 2022 at 1:54

You must log in to answer this question.