I am studying measure theory in mathematics. Let $X$ be a metric space, and $f\: X \to X$ be a continuous map. A Borel measure $\mu$ on $X$ is called $f$-invariant if $f_*\mu =\mu$, or equivalently, if $\mu(f^{-1}(B))=\mu(B)$ for all Borel sets $B\subseteq X$.
Consider two $f$-invariant Borel probability measures
$\nu$ and $\mu$ on $X$ and let $\nu=\nu_s+\nu_a$ be the Lebesgue decomposition of $\nu$ with respect to $\mu$, where
$\nu_s\perp \mu$ and $\nu_a<<\mu$. Show that then $\nu_s$ and
$\nu_a$ are also $f$-invariant.
Of course, if only one of $\nu_s$ or $\nu_a$ is $f$-invariant, then both of them are invariant. By looking at the proof of the Lebesgue-Radon-Nikodym decomposition of $\nu$, we may assume that $\nu_s$ is a positive measure and that $\nu_a = h d\mu$ where $h$ is a $\mu$-integrable function. I have proven that if $\mu(^{-1}(B)) = 0 = \mu(B)$, then both $\nu_a(f^{-1}(B))$ and $\nu_a(B)$ is zero, and hence also $\nu_s(f^{-1}(B)) = \nu_s(B) = 0$. I also have proven that $\nu_s(B) = \nu_s(f^{-1}(B \cap P))$ where $X = P \bigsqcup N$, where $N$ is $\nu_s$-null and $P$ is $\mu$-null.