if $\mu(X)$ is finite then $\lim_{n\to \infty}\mu \{x: |f(x)|\geq n\}=0$

Question

Let $(X,\mathcal F, \mu)$ be measurable space with $\mu(X)<\infty$. a) Prove that if function $f$ is measurable on $X$ then $$\lim_{n\to \infty}\mu \{x: |f(x)|\geq n\}=0.$$

b) Can we discard the hypothesis $\mu(X)<\infty$?

I tried to use Chebyshev inequality but it doesn't work. Can anyone give me some hints? Many thanks

Answer 1

We cannot discard the hypothesis of finiteness of $\mu$. With the normal Lebesgue measure and with $f(x) = x$, say, we have $\mu\{x:|x|\ge n\} = \infty$ for all $n$.

Answer 2

Let $A_n:=\{x\in X\ :\ |f(x)|\geq n\}$ for $n\in\mathbb N_0$. Then $A_{n+1}\subset A_n$ for each $n$. Now use that $\mu$ is continuous from above. Note that this requires that at least one of the sets $A_n$ has finite measure, and this is the case since $X$ has finite measure.

As pointed out in another answer, the assumption that $\mu(X)<\infty$ cannot be discarded.