# Intuition on Dominating Measures and Absolute Continuity

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.

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 $$\mu(A) < \epsilon$$.