TITLE: Proof that $dx/|x|$ is a Haar measure on non-zero reals?
QUESTION [6 upvotes]: Most importantly, what is the meaning of this notation $\lambda = dx/|x|$?  How do I compute say $\lambda(0,1)$ for example?

REPLY [10 votes]: This is the Haar measure on the multiplicative group ${\bf R}^\times$ (or the group of positive reals under multiplication too, but then there's no reason to use the absolute value symbol). Note that the "$dx$" part actually corresponds to the additive Haar measure. In symbols,
$$\mu^\times(A):=\int_Ad^\times x=\int_A\frac{d^+x}{|x|}.$$
(The $d^+x$ means we perform the integral as usual in calculus, because our measure there is in fact the additive Haar measure.)
For example, let's look at an interval $I=(a,b)$ with $0<a<b$. Then
$$\mu^\times(I)=\int_{(a,b)}d^\times x=\int_a^b\frac{dx}{|x|}=\log b-\log a=\log\left(\frac{b}{a}\right).$$
Notice, then, that if $cI=(ca,cb)$ where $c>0$ is some multiplicative scaling factor,
$$\mu^\times(cI)=\log\left(\frac{cb}{ca}\right)=\log\left(\frac{b}{a}\right)=\mu^\times(I).$$
This is an example of how the multiplicative Haar measure is translation-invariant, where translation in the multiplicative group amounts to uniform multiplication by a fixed scalar.
More generally, if $A\subset(0,\infty)$, we have
$$\mu^\times(cA)=\int_{cA}d^\times x=\int_{x\in cA}\frac{dx}{|x|}=\int_{u\in A}\frac{cdu}{|cu|}=\int_A\frac{du}{|u|}=\mu^\times(A).$$
This follows from the simple substitution $x=cu$, $dx=cdu$, $x=cu\in cA\iff u\in A$.