Prove that $\lim_{t\to \infty} t\mu(\{x:f(x)\geq t\})=0$ Problem
Suppose $f$ is a non-negative integrable function on a measure space $(X,\mathcal{A},\mu).$
Prove that $$\lim_{t\to \infty} t\mu(\{x:f(x)\geq t\})=0$$
Attempt
Let $E_t=\{x:f(x)\geq t\}$
Note that $f\chi_{E_t}\leq f$. Since $f$ is integrable, the Lebesgue dominated convergence theorem tells us that $$\lim_{t\to\infty} \int f\chi_{E_t}\rightarrow \int \lim_{t\to\infty} f\chi_{E_t}=0,$$ since $\lim_{t\to\infty} f\chi_{E_t}=0$ a.e. $\ ^{(1)}$ The result follows from the fact that $$t\mu(E_t)\leq\int f\chi_{E_t}.$$
Question
I am having trouble proving $(1)$. That is, how can I show that $f\chi_{E_t}\rightarrow 0$ a.e?
 A: Suppose $x\in X$ is such that $f(x)<\infty$. Then, for $t>f(x)$, one has that $x\notin E_t$ or $\chi_{E_t}(x)=0$. Therefore, $\lim_{t\to\infty}f(x)\chi_{E_t}(x)=0$. This shows that $$\{x\in X\,|\,f(x)<\infty\}\subseteq \left\{x\in X\,\Big|\,\lim_{t\to\infty}f(x)\chi_{E_t}(x)=0\right\}.$$ But the integrability of $f$ implies that the set $G\equiv\{x\in X\,|\,f(x)=\infty\}$ has zero measure (otherwise, $\int f\,\mathrm d\mu\geq\int_G f\,\mathrm d\mu=\mu(G)\times\infty=\infty$ if $\mu(G)>0$). It follows that the set $\{x\in X\,|\,f(x)<\infty\}$ has a negligible complement and so does the set $\{x\in X\,|\,\lim_{t\to\infty}f(x)\chi_{E_t}(x)=0\}$.
A: Recall that

$$\int_X |f|\,d\mu= \sum_{n=0}^{\infty}\int_{F_n} |f|\,d\mu$$

where $F_n =\{x\in X : n\le |f|<n+1$.
Since the series on the RHS converges, and since

$$n\mu(F_n)\le\int_{F_n}|f|\,d\mu$$

we have that
$$\lim_{n\to\infty} n\mu(F_n)\to 0.$$
However, when $n\le t<n+1$ we have
$$E_t\subseteq \bigcup_{k=n}^\infty F_n=G_n$$
But then $\mu(G_n)$ is an absolutely convergent sum of $o(n^{-1})$ functions, so is itself $o(n^{-1})$, and so

$$0\le \lim_{t\to\infty}t\mu(E_t)\le \lim_{n\to\infty} (n+1)\mu(E_t)\le\lim_{n\to\infty} (n+1)\mu(G_n)=0.$$

A: To complement all the solutions here, I'd like to add that  by applying a similar reasoning, one can show that if $f\geq0$, $\phi:[0,\infty)\rightarrow[0,\infty)$ is monotone nondecreasing, not identically $0$,  and $\phi\circ f$ is integrable, then
$$\lim_{t\rightarrow\infty}\phi(t)\,\mu(f>t)=0$$
To see this, notice that for all $t$ large enough $$\mu(f>t)\leq \mu(\phi\circ f\geq \phi(t))\leq\frac{1}{\phi(t)}\int_{\{\phi\circ f>t\}}\phi\circ f\,d\mu$$
Since $\phi\circ f\in L_1(\mu)$, $\mu(\phi\circ f=\infty)=0$ and so, by dominated convergence
$$\lim_{t\rightarrow\infty}\int_{\{\phi\circ f>t\}}\phi\circ f\,d\mu=0$$
Common examples of this slight generalization are $\phi(t)=t^p$, and $\phi(t)=e^{t}$ which lead to
$$\lim_{t\rightarrow\infty}t^p\mu(|f|>t)=0\qquad\text{if}\qquad{f\in L_p}$$
$$\lim_{t\rightarrow\infty}e^t\mu(|f|>t)=0\qquad\text{if}\qquad{e^{|f|}\in L_1}$$
