TITLE: Measure Theory - Absolute Continuity
QUESTION [8 upvotes]: A question from my homework:

Suppose $\mu$ is a measure on the real line w.r.t to the Borel $\sigma$-algebra such that $\forall x \in \mathbb{R}$ $\mu(\{x\})=0 $. is $\mu$ necessarily absolutely continuous w.r.t. to the Lebesgue measure? 

We say $\mu$ is absolutely continuous w.r.t. to $m$ if for every measurable set $E$ such that $m(E)=0$ than $\mu(E)=0$ also.
I can't see a reason why this should be true, but can't find a good counter example. I tired some variations on the counting measure. Also tried to construct some measures that give positive measure on uncountable sets and zero measure on countable, be these come out non-additive (i.e. not measures).
Would greatly appreciate the help.
Thanks!

REPLY [4 votes]: No. Every Borel measure which does not contain Dirac measures (non-atomic) and is singular with respect to the Lebesgue measure has this property.
For example the uniform probability measure on the Cantor set which is defined as follows.
Let $C$ be the Cantor set which is defined as $C=\cap_{n=1}^\infty I_n$, where $I_n$ is a union of $2^n$ disjoint closed interval, each of length $3^{-n}$. Then define the functional
$$
\ell_n(f)=\frac{3^n}{2^n}\int_{I_n}f\,dx, \quad f\in C[0,1].
$$
Clearly, $|\ell_n(f)|\le \|f\|_{\infty}$. It is not hard to show that $\{\ell_n\}$ converges weakly in ${\mathcal M}[0,1]$, and its limit $\ell$ is realized by a positive Borel measure $\nu$, i.e.,
$$
\ell(f)=\int_{[0,1]}f\,d\nu,
$$
which has the required properties.