Problem statement: Suppose that $\mu$ is a finite measure. Prove that a measurable, non-negative function $f$ is integrable if and only if $\sum^\infty_{n=1} \mu(\{x \in X : f(x) \ge n\}) < \infty$.

My attempt at a solution: Let $A_n = \{x \in X : f(x) \ge n\}$. To show that $\sum^\infty_{n=1} \mu (\{x \in X : f(x) \ge n\}) < \infty$ implies $f$ is integrable, the Borel Cantelli lemma tells us that almost all $x \in X$ belong to at most finitely many $A_n$. Thus, the set $\{x \in X : f(x) = \infty\}$ has measure $0$. Now, this, together with the fact that $\mu(X) < \infty$, should give us that $f$ is integrable, but I can't figure out how to prove that! It seems fairly obvious, but I can't figure out if it is then ok to say that $f$ is bounded almost everywhere? It seems like $f$ is then pointwise bounded, but I'm not sure if that means I can find some $M$ such that $f(x) \le M$ for all $x$.

For the reverse implication, I haven't come up with anything useful - I have been trying to show that the sequence of partial sums, $\sum^m_{n=1}\mu(A_n)$, is bounded, but I'm not sure how to do so.

  • $\begingroup$ What do you mean by your summand formula that you are summing over (with respect to $n$, in your problem statement)? Is it the measure of the set you indicated in the notation? $\endgroup$ Aug 6, 2015 at 19:45
  • $\begingroup$ @user2566092 yes, sorry, totally screwed up there! it's been edited $\endgroup$
    – poppy3345
    Aug 6, 2015 at 19:46
  • $\begingroup$ @gesa: Doesn't a convergent sum like , $\Sigma_{n=1}^{\infty} \mu(A_n) $ have bounded partial sums? $\endgroup$
    – Gary.
    Aug 6, 2015 at 20:48

3 Answers 3


Convince yourself that $$f(x)-1\leq \sum_{n=1}^\infty {\bf 1}_{(f\geq n)}(x)\leq f(x)$$ for all $x\in X$, then integrate with respect to $\mu$.

  • 1
    $\begingroup$ Very nice solution. It also shows that in one direction we don't need $\mu$ to be a finite measure. $\endgroup$
    – Ramiro
    Aug 6, 2015 at 22:25

Define $A_k=\{x:f(x)\geq k\}$ (as you had done so) and $B_k=\{x:f(x)\in[k,k+1)\}$. The $B_k$ are pair-wise disjoint. We have $\displaystyle X=\bigcup_{k=0}^\infty B_k$. Also note that $\displaystyle A_n=\bigcup_{k=n}^\infty B_k$. This gives us $\displaystyle \mu(X)=\sum_{k=0}^\infty \mu(B_k)$ and $\displaystyle \mu(A_n)=\sum_{k=n}^\infty \mu(B_k)$.

Assume non-negative $f:X\rightarrow \Bbb R$ is integrable, then

$$\infty>\int_X f d\mu \geq \sum_{k=1}^\infty k\mu(B_k)= \sum_{k=1}^\infty \mu(A_k). $$

Writing out $\mu(B_k)$ each $k$ times in a list/grid and rearranging the sum appropriately shows the last equality. This gives one direction.

For the other direction, assume $\displaystyle \sum_{k=1}^\infty \mu(A_k)<\infty$.

Since $f$ is measurable and bounded on each $B_k$, it is integrable on each $B_k$ (prove this), and we have

$$ \begin{aligned} \int_{X} f d\mu&=\lim_{N\rightarrow\infty}\sum_{k=0}^N \int_{B_k} f d\mu\\ &\leq \lim_{N\rightarrow\infty}\sum_{k=0}^N (k+1)\mu(B_k) \\ &=\mu(X)+\mu(A_1)+\mu(A_2)+\cdots<\infty \end{aligned} $$

Again, writing out $\mu(B_k)$ each $k+1$ times in a list/grid and rearranging the sum appropriately shows the last equality.

  • $\begingroup$ Why do you use $\lim\limits_{n \to \infty} \sum\limits_{k = 0}^{n}$ instead of $\sum\limits_{k = 0}^{\infty}$ in the other direction? $\endgroup$
    – Ramanujan
    Jan 23, 2019 at 19:22

If $f\ge 0$, you have $$\int_\Omega f\, d\mu = \int_0^\infty |[f \geq x]| \, dx.$$ The function $x\mapsto |[f \geq x]|$ decreases to $0$ at $\infty$. Can you see the rest?

  • $\begingroup$ I use $|E|$ for the Lebesgue measure of $E$. $\endgroup$ Aug 6, 2015 at 19:28
  • $\begingroup$ Is this saying that $\int f = \int m(\{f(x) : f(x) \ge x\})$? @ncmathsadist $\endgroup$
    – poppy3345
    Aug 6, 2015 at 20:10

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