# Prove that $f$ is integrable if and only if $\sum^\infty_{n=1} \mu(\{x \in X : f(x) \ge n\}) < \infty$

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.

• 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? Commented Aug 6, 2015 at 19:45
• @user2566092 yes, sorry, totally screwed up there! it's been edited Commented Aug 6, 2015 at 19:46
• @gesa: Doesn't a convergent sum like , $\Sigma_{n=1}^{\infty} \mu(A_n)$ have bounded partial sums? Commented Aug 6, 2015 at 20:48

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 $\mu$.

• Very nice solution. It also shows that in one direction we don't need $\mu$ to be a finite measure. Commented 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.

• 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? Commented Jan 23, 2019 at 19:22
• I believe that, generally, $\sum_{k=0}^\infty$ is defined to actually mean $\lim_{n\to\infty}\sum_{k=0}^n$. So the choice of writing one or the other can be a matter of personal preference. Sometimes it is important to write the finite sum first before taking the limit, but I don't think that's an issue here. Commented May 5 at 18:24

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?

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

Note that if $$f$$ is non-negative and integrable, we have: $$\lim_{t \rightarrow \infty} t \cdot \mu(\{x \in X : f(x) \geq t \}) = 0$$ The above can be proved with Markov's inequality and the Dominated Convergence Theorem. This implies then that for large enough $$N$$, there exists an $$\epsilon > 0$$ such that $$\mu(\{x \in X : f(x) \geq t \}) \leq \frac{1}{t^{1+\epsilon}}$$ for all $$t \geq N$$. Therefore, $$\sum \mu(\{x \in X : f(x) \geq n \})$$ is bounded by a $$p$$-series, and hence converges.