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?
Best Answer
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 $\sup_n(-\int f_n)=-\inf_n\int f_n$.