Absolutely continuous invariant measures for a minimal flow Fix $(X, g)$ to be some compact Riemannian manifold and $V \in \Gamma(TX)$ a smooth, non-vanishing vector field. Suppose the flow is minimal, i.e. every orbit is dense in $X$, and volume-preserving. 
Suppose there exists a measure $\sigma$ that is both invariant under the flow of $V$ and absolutely continuous with respect to the Riemannian volume measure. In other words, it is given by integration against $f \text{dvol}_g$ for $f \in L^1(X)$. 
If $f$ is not merely $L^1$ but continuous, then our topological assumptions imply immediately that $f$ is constant, so actually any invariant measures are a constant multiple of the volume measure (the minimality precludes the existence of any singular invariant measures). 
Can we construct an example such that $f$ is not constant almost everywhere? 
 A: It seems to me that the answer to your question is "yes", and in fact,
Furstenberg's example can be used to construct such an example.
The idea of Furstenberg's construction in the paper Strict Ergodicity and Transformation of the Torus consists in finding an
irrational number $\alpha$ and a $\phi\in
C^\infty(\mathbb{T}^1,\mathbb{R})$ with $\int\phi(x) d\mathrm{Leb}(x) = 0$ such that there is a
measurable $L^1$ function $u\colon \mathbb{T}\to\mathbb{R}$ satisfying
$\phi(x) = u(x+\alpha) - u(x)$ for Lebesgue a.e. $x\in\mathbb{T}$, but
there is no continuous function $u$ satisfying this property.
Then, the diffeomorphism $g : \mathbb{T}^2 \ni (x,y) \mapsto
(x+\alpha,y+\phi(x))$ is minimal, preserves the Lebesgue measure of
$\mathbb{T}^2$, but this is not an ergodic measure. In fact, the map
$h : (x,y)\mapsto (x,y+u(x))$ leaves invariant the Lebesgue measure of
$\mathbb{T}^2$ and $g\circ h = h \circ k$, where $k : (x,y)\mapsto
(x+\alpha, y)$; and Lebesgue is clearly not ergodic for $k$.
However, there are infinitely many $k$-invariant measures which are
absolutely continuous with respect to Lebesgue. To see this, one can
consider for instance a non trivial interval $I\subset\mathbb{T}$
(i.e. $I$ and its complement contain more than one point), and
define the measure $\mu:= \mathrm{Leb} \otimes
\frac{1}{\mathrm{Leb}(I)}\mathrm{Leb}\big|_I$. Then, since $h$
preserves Lebesgue, we conclude that $h_*\mu$ is $g$ invariant, is
absolutely continuous with respect to Lebesgue and is different from
Lebesgue. 
