Showing $t\mu(|f|>t) \to 0$ as $t \to \infty$ I am working on some qual problems on Analysis, here is one I want to ask.
As in the title, let $(X,\mu,\mathcal{M})$ be a measure space, let $f \in L^1(\mu)$, then we want to show the limit $t\mu(|f|>t) \to 0$ as $t \to \infty$.
My approach is to first observe that if $f \in L^2$ then this would hold by the Chebyshev inequality, then by noting $L^1 \cap L^2$ is dense in $L^1$, I can do some approximation argument then. Would it work?
 A: Assume that $t\mu\{|f|>t\}$ doesn't tend to $0$, we'll show that $f$ cannot be integrable. By assumption, there's a $\delta>0$ and a sequence $\{t_n\}$ tending to infinity such that $t_n\mu(A_n)>\delta$, where $A_n = \{x\colon |f(x)|>t_n\}$. If $\mu(A_n)$ is bounded away from $0$, then it's clear that $f$ is not integrable, since $\int_X |f|\,d\mu \ge \int_{A_n} t_n\,du$ which would tend to infinity. Otherwise, by replacing $\{t_n\}$ by a subsequence, we may assume that $\mu(A_n)$ tends to $0$, and in fact that $\mu(A_{n+1})<\frac12\mu(A_n)$.
Note that $A_1 \supset A_2 \supset \cdots$, and so the sets $B_n = A_n \setminus A_{n-1}$ are disjoint; note also that $\mu(B_n) = \mu(A_n) - \mu(A_{n-1}) > \frac12\mu(A_n)$. Then
$$
\int_X |f|\,d\mu \ge \sum_{n=2}^\infty \int_{B_n} |f|\,du \ge \sum_{n=2}^\infty \mu(B_n) \inf_{x\in B_n} |f(x)| > \sum_{n=2}^\infty \tfrac12\mu(A_n) t_n > \sum_{n=2}^\infty \tfrac\delta2 = \infty
$$
as claimed.
A: Let $A_t=\{x\in X:|f(x)|>t\}$, then $A_{t+1}\subset A_t$. Let $f_t=f\chi_{A_t}$, then since $f\in L^1(X,m)$, $|f|$ is finite a.e. and hence $f_t\to 0$ a.e. as $t\to \infty$. Also we have $|f_t|\leq |f|$, so by DCT
\begin{align*}
0=\lim_{t\to\infty} \int_X f\chi_{A_t}\,d\mu\geq \lim_{t\to\infty} \int_X n\chi_{A_t}\,d\mu=\lim_{t\to\infty}  n \,\mu(A_n).
\end{align*}
