Let $E$ be a measurable subset of $\mathbb{R}$ and $f \in L^1(E)$. Set $A_n=\{ x \in E: |f(x)| \geq n \}$ for all $n \in \mathbb{N}$. Prove that $$\sum^{\infty}_{n=1} m(A_n) \leq \int_{E}|f|.$$

My Attempt:

By Chebychev's Inequality for each $n$ we have $$m(A_n) \leq \frac{1}{n} \int_E|f|$$ But when I add all such inequalities I'd get $$\sum^{\infty}_{n=1} m(A_n) \leq \sum^{\infty}_{n=1} (\frac{1}{n} \int_{E}|f|)$$ Since $f$ is integrable we can write $$\sum^{\infty}_{n=1} m(A_n) \leq \left[ \int_{E}|f| \right]\sum^{\infty}_{n=1} (\frac{1}{n} )$$ But the problem is that the series diverges (let alone that I need it to be precisely equal to 1!)

Thank you for your hints and ideas


Indeed, the use of Chebychev is too crude.

Define $B_k:=A_k\setminus A_{k+1}$; we have $\int_{B_k}|f|\mathrm d\mu\geqslant k\mu(B_k)$. Since the family $(B_k)$ is pairwise disjoint, we have $\sum_k k\mu(B_k)\leqslant \int_X|f|\mathrm d\mu$. Conclude using summation by parts.


Here's an outline a possible proof.

  1. show that left hand side equals the measure of a set S in R^2: S = {(x, y) | x in E and |y| < floor(|f(x)|)}

  2. show that right hand side equals the measure of a set T in R^2: T = {(x, y) | x in E and |y| < |f(x)|}

  3. since S is contained in T the inequality we want to prove follows from monotonicity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.