Riesz representation theorem:
Let $X$ be a locally compact Hausdorff space, and let $\Lambda$ be a positive linear functional on $C_c(X)$. There exists a $\sigma$-algebra $\mathfrak{M}$ in $X$ which contains all Borel sets in $X$, and there exists a positive measure $\mu$ on $\mathfrak{M}$, s.t. $$\Lambda(f)=\int_X f d\mu \; \; \; \; \forall f \in C_c(X)$$
with the additional properties:
(i) $\mu(K)<\infty$ for every compact set $K\subset X$
(ii) For every $E\in \mathfrak{M}$, we have $$\mu(E)=\inf\lbrace \mu(V) : E\subset V,\; \; V \; \mbox{open} \rbrace$$
(iii) The relation
$$\mu(E) = \sup\lbrace \mu(K) : K\subset E,\; \; K \; \mbox{compact} \rbrace $$
holds for every open set $E$, and for every $E\in \mathfrak{M}$ with $\mu(E)<\infty$.
(iv) If $E\in \mathfrak{M}$, $A\subset E$, and $\mu(E)=0$, then $A\in \mathfrak{M}$.
Question:
Given a positive linear functional $\Lambda$ on $C_c(\mathbb{R})$ in which cases the measure $\mu$, obtained as in the Riesz Representation Theorem, is absolutely continuos with respect to the Lebesgue measure on $\mathbb{R}$?
No.
Take $X=\Bbb R$. The classical counterexample is the Dirac measure $\delta\in C_c(\Bbb R)^*$ defined by $$ \delta(f) := f(0). $$ It is easy to see that $\delta$ is indeed a continuous linear functional. However, it corresponds to an atomic measure $$\delta(A)=\begin{cases} 1 &; 0\in A\\ 0 &;0\ne A\end{cases}.$$ This measure is not absolutely continuous with respect to the Lebesgue measure.