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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.

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  • $\begingroup$ 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. $\endgroup$
    – rod
    Jan 11 at 16:54

1 Answer 1

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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.

It follows that any measure is absolutely continuous with respect to counting measure, while very few functions are integrable.

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  • $\begingroup$ 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. $\endgroup$
    – Jacob
    Dec 1, 2014 at 5:16

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