Let us consider $T:[0,1]\rightarrow [0,1]$ be a continuous map. Let $\mu, \nu$ be $T$ invariant probability measures on $\mathcal{B}([0,1])$. I want to show that if there exists $D>0$ s.t. $\int_{[0,1]} f~d\nu\leq C\int_{[0,1]} f~d\mu$ for all $f:[0,1]\rightarrow \Bbb{R_+}$ continuous then $\nu \ll \mu$ (absolutely continuous).
My idea was first to consider some simple functions $f=1_{[a,b]}$ where $[a,b]\subset [0,1]$. Then assume that there exists $C$ s.t. $\nu([a,b])\leq C\mu ([a,b])$. Now I need to pick an arbitrary $A\in \mathcal{B}([0,1])$ and assume $\mu(A)=0$. I want to show that $\nu(A)=0$. I thought that since sets of the form $[a,b]$ generates $\mathcal{B}([0,1])$ we know that $A$ can be written as unions and intersections of intervalls $[a,b]$. But I'm not really sure if this can be done like this. Because I would then try to use operations of the measure on unions and intersections to conclude that $\nu(A)\leq C\mu(A)$ which allows us to conclude in this case if we assume $\mu(A)=0$.
But does this work?
If $K$ is a compact subset of $[0,1]$ then there is a decreasing sequence $(f_n)$ of continuous functions with pointwise limit $1_K$ (the indicator of $K$). This shows $\nu(K)\le C\mu(K)$ for all compact $K$. Now use inner regularity of $\nu$.