Let $(X,B,\mu)$ be a measure space and $T:X\to X$ be a transformation such that $\mu$ is ergodic w.r.t. $T$. See the Wikipedia article for further details of the ergodicity setting. I am looking to understand the claim that if $\nu$ is any measure which is absolutely continuous with respect to a probability measure $\mu$, then $\nu$ is a constant multiple of $\mu$.
Currently I am stuck at using the assumed invariance/ergodicity of $\mu$. To be precise, if $\nu\ll\mu$ then the Radon-Nikodym theorem guarantees the existence of $f\in L^1(\mu)$ such that $\nu(A) = \int_Afd\mu$ for any $\nu$-measurable set $A$. Now we ought to show that $f = C$ $\mu$-a.e. in $X$. Since I don't know what such a constant $C$ might be my idea was to use contradiction: So if $f$ is not constant $\mu$-a.e. there exists some $N\in B$ such that $f$ is not constant on $N$ and $\mu(N) > 0$. By assumption $\mu(N) = \mu(T^{-1}(N))$ and this is probably the part where the contradiction assumption should kick in: I would need to conclude that somehow $\mu(N)\neq \mu(T^{-1}(N))$ by using the definition of $\nu$. But
$$\nu(N) = \int_Nf(x)d\mu(x) = \int_X 1_N(x)f(x)d\mu(x)$$
and
$$\nu(T^{-1}(N)) = \int_{T^{-1}(N)}f(x)d\mu(x) = \int_X1_N(T(x))f(x)d\mu(x)$$
and I don't see a priori anything connecting $\nu(N)$ to $\nu(T^{-1}(N))$. What should I do?