real analysis, chebyshev's inequality Suppose $f$ is a non negative integrable function on a measure space $(X,M,μ)$. 
Prove that:
$$\lim_{t \rightarrow \infty} t\cdot \mu(\{x:f(x)\geq t\} )=0.$$ 
Can you help me please?
 A: Hint:
$$
f(x) 1_{f(x)\ge t}\ge t 1_{f(x)\ge t}
$$
details:
$$t\mu\{x:f(x)\ge t\} = \mu(t 1_{f(.)\ge t})
\le \mu(f 1_{f(.)\ge t})=
\int_t^\infty \mu\{f(.)\ge x\}dx
$$Now use the dominated convergence theorem.
A: $$\lim_t t\mu(\{f\geq t\})\leq \lim_t \int_X f \chi_{\{f\geq t\}} d\mu = \int_X f \chi_{\{f = \infty\}} d\mu = 0$$
since $\{f=\infty\}$ has to be a null set. Passing of the limit is done by dominated convergence theorem with the bounding function $f$.
A: Suppose this isn't true. Then $\exists \epsilon >0$ and $\exists (t_n) \subset \mathbb{R}$ such that $\lim_{n \rightarrow \infty} t_n = \infty$ such that 
$$\forall n \in \mathbb{N} \quad  \mu \{x: f(x) \geq t_n\} > \dfrac{\epsilon}{t_n}.$$
Let $A_n := \{x: f(x) \geq t_n\}$. By standard arguments we can assume that these sets are disjoint and nonempty. But then
$$\int f d\mu \geq \sum_n \dfrac{\epsilon}{t_n} \cdot t_n = \sum_n \epsilon = \infty,$$
which is a contradiciton. 
A: You have, for all $t\geq 0$ $$0 \leq t\mu\{x\in X \colon f(x)\geq t\} \leq \int_X f(x)\mathbf{1}_{\{f(x)\geq t\}} \mu(dx).$$
Let $h(t,x)\stackrel{\rm def}{=}f(x)\mathbf{1}_{\{f(x)\geq t\}}$.


*

*For any fixed $x\in X$,  $h(x,t)\xrightarrow[t\to\infty]{} 0$;

*$0 \leq h(x,t) \leq f(x)$ for all $(x,t)$;

*$\int_x f < \infty$.


Thus, using (for instance; monotone convergence would also work) the dominated convergence theorem, $$\int_x h(t,x) \mu(dx) \xrightarrow[t\to\infty]{} 0.$$
A: Let $g$ be a simple function such that $g\leqslant f$. For each $t$, 
$$t\mu\{f\geqslant t\}\leqslant t\mu\{f-g\geqslant t/2\}+t\mu\{g\geqslant t\}\leqslant 2\int |f-g|\mathrm d\mu+t\mu\{g\geqslant t\}.$$
Since $g$ is bounded, we obtain 
$$\limsup_{t\to +\infty}t\mu\{f\geqslant t\}\leqslant 2\int |f-g|\mathrm d\mu.$$
As $g$ was arbitrary, we conclude by definition of Lebesgue integral. 
