I'm trying to find an example to show that the following theorem doesn't hold if $\nu$ is infinite:

Let $\nu$ be a finite signed measure and $\mu$ a positive measure on $(X, \mathcal{M})$. Then $\nu \ll \mu$ ($\mu E = 0 \Rightarrow \nu E = 0$) iff for every $\epsilon > 0$, there exists $\delta > 0$ such that $|\nu(E)| < \epsilon$ whenever $\mu(E) < \delta$.

Based on the proof of this theorem, I think my best bet is to find a pair of measures such that $\nu \ll \mu$, but the $\epsilon-\delta$ condition doesn't hold. i.e. a pair of measures such that every set of $\mu$-measure $0$ has $\nu$-measure $0$, but there is some $\epsilon$ such that we can find arbitrarily $\mu$-small sets of $\nu$-measure at least $\epsilon$.

What types of measures should I be looking for? I don't really have any intuition yet for signed measures. Whenever I had to find examples using positive measures, I could usually do it with variations on Lebesgue measure and integrals of the dirac delta function, but that seems fruitless here.

  • $\begingroup$ Just take $\mathbb R$, $\mu = dx$ the Lebesque measure and $\nu = x^2 dx$. $\endgroup$ – user99914 Nov 17 '14 at 6:45

To avoid over complicating things consider $\nu$ a $\sigma$-finite measure. Then since we want $\nu \ll \mu$ the Radon-Nykodim theorem says that $d\nu = fd\mu$ for some non-negative $\mu$-measurable $f$. Now you want to look for $f$'s that are very big on small sets of $\mu$. Two examples:

  1. Consider $X=\mathbb{N}$ and $\mu(A)= \sum_{j\in A} 2^{-j}$, $\nu(A)= \sum_{j\in A} 2^j$. Then what can you say about $\mu(\{ n\})$ and $\nu(\{ n\})$ as $n\to \infty$?

  2. Consider $X=\mathbb{R}$, $\mu$ Lebesgue measure and $d\nu=e^x d\mu$, then what can you say about $\nu([n, n+1/n])$?


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