Comparing lebesgue measure and counting measure

I have the following problem from Folland:

Let $X = [0, 1]$, $\mathcal{M} = \mathcal{B}_{[0, 1]}$, $m =$ Lebesgue measure and $\mu =$ counting measure.

1. $m \ll \mu$ but $dm \neq f \, d\mu$ for any $f$.
2. $\mu$ has no Lebesgue decomposition with respect to $m$.

I think I might be understanding counting measure incorrectly, because it seems to me that $m \ll \mu$ is not true, because any Borel subset of the open interval $(0, 1)$ would have counting measure zero because it does not contain any integers, but clearly could have positive Lebesgue measure.

• Note for my future self: although the Lebesgue measure is AC w.r.t. the counting measure, it has no density in terms of the latter. This is because the Radon-Nikodym theorem holds only for sigma-finite measures (the space has to be countable union of finite-measure sets). And the counting measure is clearly not sigma-finite.
– rod
Jan 11 at 16:54

The counting measure does exactly what it says: it counts the number of elements in a set. So any infinite set has counting measure $\infty$, while the measure of any finite set is its cardinality. It has nothing to do with integers in particular.
• Okay, this makes more sense. I've seen counting measure on $\mathbb{R}$ used to mean $\mu(E) = |E \cap \mathbb{Q}|$, but I guess that was an abuse of terminology. Dec 1, 2014 at 5:16