Show $f$ : integrable $\Leftrightarrow$ $ \sum_{n=1}^{\infty} \mu\left( \{ x \in E : |f(x)| \ge n \} \right) < \infty $

Question

Let $f : E \to [0, \infty]$ measurable, where $E$ is a finite measure space.

Show : $f$ is integrable if and only if

$$ \sum_{n=1}^{\infty} \mu\left( \{ x \in E : |f(x)| \ge n \} \right) < \infty $$

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Try

($\Rightarrow$)

Let $E_n = \{ x \in E : |f(x)| \ge n \}$.

Since $E_1 \supset E_2 \supset \cdots $, we have

$$ E = (E\setminus E_1) \sqcup (E_1 \setminus E_2) \sqcup \cdots $$

Therefore, $\int_E f d\mu = \int_{E\setminus E_1} f d\mu + \int_{E_1\setminus E_2} f d\mu + \cdots$

But I cannot proceed from here.

($\Leftarrow$)

We have $\sum_{n=1}^{\infty} \mu\left( E_n \right) = \sum_{n=1}^{\infty} \int \chi_{E_n} d\mu$,

By Levi's convergence theorem, $\sum_{n=1}^{\infty} \chi_{E_n}$ converges a.e. in $E$.

But I'm stuck at how I can relate this to the integrability of $f$.

Answer 1

Hint: Note that $|f| \ge \sum_{n \ge 1} \chi_{E_n}$ pointwise.

So if $f$ is integrable, then $$\sum_{n \ge 1} \mu(E_n) = \sum_{n \ge 1} \int \chi_{E_n} \, d\mu = \int \sum_{n \ge 1} \chi_{E_n} \, d\mu \le \int |f| \, d\mu < \infty$$ where the interchange of integral and sum is due to the monotone convergence theorem.