I have to show that

$\lim_{n \to \infty} n\cdot m(\{ x \in A | |f(x)| \geq n\}) = 0$, for $(A, \textit{S}, m)$ a measure space and $f: A \rightarrow \mathbb{R}$ an integrable function.

Let $A_n = \{ x \in A | |f(x)| \geq n\}$

Here's what I've been thinking:

I know that $\forall n >0: n\cdot m(A_n) \leq \int_{A_n}|f(x)| dm(x)$ so I thought that this might mean that I have to show that $f(x) = 0$ so that $\int f = 0$, but this seemed odd to show that for a general function. I also noticed, that the $A_n$ form a decreasing sequence, i.e. $A_{n+1} \subset A_n$, which means that $m(\bigcap_{n\geq 0} A_n) = lim_{n\to \infty} m(A_n)$. However I don't know if this knowledge is useful or how to apply it.

If someone could point me in the right direction I'd really appreciate it. Thanks!


1 Answer 1


A corretion there, it should be $m\left( \cap_n A_n\right) =\lim_n m(A_n)$, since the sequence decreases, but for this you require $m(A_n)<\infty$ for some $n$ (check this). Now note that your function must be finite almost everywhere, else it'd not be integrable.

Finally, don't forget $\mu(B) = \int_B |f|dm$ is a measure too , what happens when $m(B) =0$ to $\mu(B)$? ;)

  • $\begingroup$ Ok, so I know there is an $n \in \mathbb{N}$ s.t. $m(A_n) < \infty$ because $m(A_n) \leq \int |f(x)| dx < \infty$ since $f$ is integrable with respect to $m$. I don't quite understand your last statement, could you elaborate please? $\endgroup$
    – Longeyes
    Commented Oct 30, 2013 at 16:24
  • $\begingroup$ @Longeyes Well, since $\mu$ is a measure, this means that $\int_{\cap_n A_n} |f| dm = \mu(\cap_n A_n) = \lim_n \mu(A_n)$ since again $\mu(A_1)\leq \int_A |f| dm < \infty$. Now, what is $m(\cap_n A_n)$ ? does this tell you anything about $\int_{\cap_n A_n} |f| dm$ ? $\endgroup$ Commented Oct 30, 2013 at 16:28
  • $\begingroup$ is $m(\bigcap_n A_n) = \mu(\bigcap_n A_n)$? $\endgroup$
    – Longeyes
    Commented Oct 30, 2013 at 16:44
  • $\begingroup$ @Longeyes What is $m(\cap_n A_n)$ (what is $\cap_n A_n$) ? Remember then that we are writing $\mu(B) = \int_B |f| dm$ $\endgroup$ Commented Oct 30, 2013 at 16:55
  • $\begingroup$ Because $A_{n+1} \subset A_n$ I thought that $\bigcap_n A_n = A_\infty$ which would make $m(\bigcap_n A_n) = m(A_\infty)$ and then...because $f$ is finite maybe $m(A_\infty) = 0$? $\endgroup$
    – Longeyes
    Commented Oct 30, 2013 at 17:05

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