Show that the measure $h(E)= \int_{E} \frac{1}{x} dx$ on Borel sets is invariant under dilations

Question

Let $E\subset(0, \infty)$ be a Borel set and measure $h(E)= \int_{E} \frac{1}{x} dx$ where the integral is Lebesgue. Given $a \in (0,\infty)$, show that

$h(E)=h(aE)$

First, note that that $h(aE)$ and $h(E)$ agree when $E$ is an interval. Also, I know that this is a Haar measure (https://en.wikipedia.org/wiki/Haar_measure) so that it is left-translation-invariant. Is that right? How I can prove that?

Answer 1

It seems all you need to do is to use the change of variable formula, $x=at$: $$h(aE) = \int_{aE}\frac{1}{x}\,dx =\int_E \frac{1}{at}\,a\,dt = h(E)$$ If you are unsure about change of variables with arbitrary Borel set present, write the integral in the (logically equivalent) way $$h(aE) = \int_0^\infty \frac{1}{x}\chi_{aE}(x) \,dx =\int_0^\infty \frac{1}{at}\chi_{aE}(at)\,a\,dt = \int_0^\infty \frac{1}{t}\chi_{E}(t)\,dt = h(E)$$