What is important about absolutely continuous invariant measures? When looking at papers, it seems that a common problem in ergodic theory is proving the existence of "absolutely continuous invariant measures". Why are such measures important? I am specifically interested in the "absolutely continuous" side of things, since I am quite comfortable with the importance of invariant measures.
Thanks!
 A: You probably mean "absolutely continuous invariant measures with respect to Lebesgue". I will assume in my answer that this is the case.
Two different but also very important reasons:


*

*From the mathematical point of view (and these measures are crucial in ergodic theory), it is quite nice that somehow they have the best possible properties.


*From the physical point of view (and these measures are crucial in describing physical phenomena), it is quite nice that sets of positive measure have positive Lebesgue measure.

In "best possible properties" I am including good ergodic properties, say more than ergodicicity, such as mixing or exponential decay of correlations (but often more). The best example, which unfortunately would require a quite lengthy discussion, is "smooth ergodic theory", where absolute continuity is crucial.
A: Consider the dynamical system $f:[0,1]\to [0,1]$ defined by $f(x)=2x\mod 1$. $f(0,1/2)=(0,1)$ and $f(1/2,1)=(0,1)$ i.e it is expanding everywhere except at zero which is a fixed point.
Invariant but not Abs Cont:
The dirac measure at zero is $f-$ invariant, however it only describe a single point and thus irrelevant from physical point of view.
Lebesgue (best case of Abs Cont):
Lebesgue measure is invariant and gives information about all the points in the space.
A: First, a short answer:

If a measure $\nu$ is absolutely continuous with respect to another measure $\mu$, then $\nu$ inherits from $\mu$ every property that holds almost everywhere with respect to $\mu$. Since invariant measures have many properties that hold almost everywhere (such as Poincaré recurrence, ergodicity and many more), it is extremely useful to find an absolutely continuous invariant measure.

Let me explain this in more details using a fundamental example. One of the most fundamental properties of invariant probability measures is the Poincaré recurrence:

Suppose that $\mu$ is a probability measure on $X$ that is invariant to a transformation $T:X\to X$. For a positive measure set $E\subset X$ we consider the set
$$F:=\left\{ x\in E:\#\left\{ n\geq1:T^{n}x\in E\right\} =\infty\right\}.$$
That is, $F$ is the set of all points $x$ in $E$ for which the orbit $x,Tx,T^2x,T^3x,\dotsc$ visits $E$ infinitely many times. Then the Poincaré recurrence theorem ensures that $\mu\left(E\backslash F\right)=0$. That is, from the point of view of $\mu$ we essentially have that "$F=E$" (more formally, the indicators $1_F$ and $1_E$ are equal as functions in $L^1\left(\mu\right)$).

Now suppose that we are given a measure $\nu$ on $X$ that is different from $\mu$ (possibly $\nu$ is not even a probability measure), and also $\nu$ is not invariant to $T$. If we can find a probability invariant measure $\mu$ such that $\nu$ is equivalent to $\mu$, then we immediately get that $\nu\left(E\backslash F\right)=0$, so that also from the $\nu$ point of view "$F=E$". Thus, we are getting the Poincaré recurrence theorem for $\nu$ "for free" (up to finding $\mu$), although this theorem concerns only invariant measures.
