Reasoning about absolute continuity between two measures by reasoning about their supports Suppose we have two measures $\mu, \nu$ on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ with $\text{supp}(\mu) \subseteq \text{supp}(\nu)$, where $\text{supp}(\mu)$ denotes the support of $\mu$, i.e.
$$
\text{supp}(\mu) := \{x \in \mathbb{R} : \forall \epsilon > 0, \mu(B(x, \epsilon)) > 0\}
$$
(where $B(x, \epsilon)$ denotes a ball of radius $\epsilon$ at $x$), and similarly for $\nu$.
I want to know when this allows us to infer that $\mu \ll \nu$. I know that we can't conclude this in general, since, for example, the hypotheses are satisfied if $\mu$ is the Dirac measure at $0$ and $\mu$ is the Lebesgue measure, and yet $\mu(\{0\}) = 1$ with $\nu(\{0\}) = 0$.
However, I wonder if we can conclude that $\mu \ll \nu$ when $\mu$ and $\nu$  have densities $f$ and $g$ respectively with respect to the Lebesgue measure? If not, does adding the assumption that $\mu$ and $\nu$ are both probability measures make this statement true?
 A: Let $A$ be a "fat Cantor set": a compact set of positive Lebesgue measure (let's say measure $1$) that is nowhere dense. 
Enumerate the intervals with rational endpoints:
$(a_n, b_n)$ where $a_n < b_n$.
 Construct inductively two sequences $B_n$ and $C_n$ of compact sets that are
affine images of $A$ ($B_n = s_n A + t_n$ and $C_n = u_n A + v_n$ with 
$s_n, u_n > 0$ and $t_n, v_n$ real), such that 


*

*$B_n$ and $C_n$ are both in the interval $(a_n, b_n)$

*These are all disjoint ($B_i \cap B_j = \emptyset$ for $i \ne j$,
$C_i \cap C_j = \emptyset$ for $i \ne j$, $B_i \cap C_j = \emptyset$ for all $i$ and $j$).


Then we can construct probability measures $\mu$ and $\nu$ such that $\mu$ is the sum over $n$ of suitable positive multiples of Lebesgue measure restricted to $B_n$,
and $\nu$ is the sum over $n$ of suitable positive multiples of Lebesgue measure restricted to $C_n$. 
The support of each is all of $\mathbb R$, because every interval with rational endpoints contains a $B_n$ whose $\mu$-measure is positive and a $C_n$ whose $\nu$-measure is positive. Each has a density with respect to Lebesgue measure, which is constant on each $B_n$ or $C_n$.
 But they are mutually singular: $\mu(\bigcup_n C_n) = \nu(\bigcup_n B_n) = 0$ while $\mu(\bigcup_n B_n) = \nu(\bigcup_n C_n) = 1$.
