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

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

Show : $$f$$ is integrable if and only if

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

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 $$f$$.

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.