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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

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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.

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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.

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