Problem 3

Exercise 7.3 (Slicing the range of an integral)   Let $X$ be a finite measure space. Let $f : X \to \mathbb{R}$ be a measurable function and define for each $k=1,2,\dots$

$$E_k = \{x \in X \;\vert\; k\leq \vert f(x)\vert < k+1\}$$ (7.6)

Then $f\in L^1(X)$ if and only if

$$\sum_{k=1}^\infty k\mu(E_k) < \infty$$ (7.7)

Proof. In the forward direction, suppose $f\in L^1(X)$. The set

$$X = \bigcup_{k=1}^\infty E_k$$ (7.8)

is always a disjoint union and we have

$$\sum_{k=1}^\infty k\mu(E_k) \leq \sum_{k=1}^\infty \int_{E_k}\vert f\vert = \int_X \vert f\vert < \infty$$ (7.9)

In the reverse direction, suppose the sum is finite. Then add $\mu(X)<\infty$ to the sum

$$\mu(X) + \sum_{k=1}^\infty k\mu(E_k) < \infty$$ (7.10)

We are free to measure the set $X$ as follows

$$\mu(X) = \sum_{k=1}^\infty \mu(E_k)$$ (7.11)

Combining the sums shows

$$\sum_{k=1}^\infty (k+1)\mu(E_k) < \infty$$ (7.12)

Compare this to the integral

$$\int_X \vert f\vert = \sum_{k=1}^\infty \int_{E_k} \vert f\vert \leq \sum_{k=1}^\infty \int_{E_k} (k+1) =\sum_{k=1}^\infty (k+1)\mu(E_k) < \infty$$ (7.13)

To see that the finite measure hypothesis is necessary, consider the function $f : \mathbb{R} \to \mathbb{R}$ sending $1\leq x\mapsto 1/x$ and $1>x\mapsto 0$. $E_k=\emptyset$ for all $k\geq 1$. Then

$$\sum_{k=1}^\infty k\mu(E_k) = 0 < \infty$$ (7.14)

But

$$\int_{-\infty}^\infty f = \int_1^\infty \frac{1}{x} = \infty$$ (7.15)

$\qedsymbol$