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From Wikipedia,

A measure $\mu$ on Borel subsets of the real line is absolutely continuous with respect to Lebesgue measure $ \lambda$ (in other words, dominated by $\lambda$ ) if for every measurable set $ A$, $\lambda (A)=0$ implies $\mu(A)=0$ . This is written as $\mu \ll\lambda$.

I understand mathematically what is written, and perhaps what is meant by it on some level. However, I cannot fully understand all the intuition behind this. Namely,

  • Why does "$\lambda (A)=0$ implies $\mu(A)=0$" mean absolute continuity. As in, is this the same continuity in the traditional $\epsilon$ - $\delta$ sense? Are they equivalent? If so .... why? The definition of "$\lambda (A)=0$ implies $\mu(A)=0$" seems so minimal to me to imply an $\epsilon$ - $\delta$ continuity
  • What is then the significance of absolute continuity of measures? Why is it important to have dominating measures, as far as probability/statistics is concerned? Is there some intuition as to why they should be defined/exist?

The original question which spurred this investigation is along the lines of:

Suppose $X$ is a random variable, and $\exists F: T = F(X)$, then the joint distribution of ($X,T$) is not dominated by a product measure.

I'm not sure what the above should mean, or why it is overall important to ensure dominance by a measure, so care is needed when defining the joint and conditional densities.

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The importance of absolute continuity comes from the Radon-Nikodym theorem. It states if $\mu \ll \nu$ then $\mu$ has a density function with respect $\nu$. A lot of the time it's much easier to prove things if all the distributions we work with have a density function, i.e. are dominated by some common measure.

It's also possible phrase absolute continuity in an $\epsilon$-$\delta$ way. It turns out $\mu$ is absolutely continuous with respect to $\nu$ if and only if for all $\epsilon > 0$ there exists $\delta > 0$ such that for all measurable $A$, if $\nu(A) < \delta$ then $\mu(A) < \epsilon$.

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  • $\begingroup$ Thanks for your reply. I added a statement onto my question by the time you answered. If possible could you also provide a reply to that statement? It is just an example I came across. $\endgroup$ Aug 15, 2019 at 14:06

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