Problem 4

Exercise 7.4 (Logarithmic Fubini FIXME) Let $f : [0,1]\to\mathbb{R}$ be integrable and set

$\displaystyle g(x) := \int_x^1 \frac{f(t)}{t}dt$

(7.16)

Proof. Then $g$

is integrable and

$\displaystyle \int_0^1 g(x)dx$	$\displaystyle = \int_0^1\int_x^1 f(t)dtdx = \int_0^1\int_0^1 \frac{f(t)}{t}\cdot 1_{x\leq t\leq 1}(t) dtdx$	(7.17)
	$\displaystyle = \int_0^1\int_0^1 \frac{f(t)}{t}\cdot 1_{0\leq x\leq t}(x) dxdt =\int_0^1\int_0^t \frac{f(t)}{t} dxdt = \int_0^1 f(t) dt$	(7.18)

$\qedsymbol$