Let $M$ be a compact metric space, and let $T$ be a homeomorphism $T \colon M \to M$.
We say measure $\mu$ on Borel $\sigma$-algebra is $T$-invariant if for every continuous function $f \colon M \to \mathbb{R}$, $\int_M f \circ T d\mu = \int_M f d\mu$.
Now let $\mu \colon M \to [0,1]$ be a $T$-invariant measure and fix Borel measurable $A$.
I want to show that $\mu_0(B) = \frac{\mu(B \setminus A)}{1-\mu(A)}$ and $\mu_1(B) = \frac{\mu(A \cap B)}{\mu(B)}$ are $T$-invariant.
It seems it would suffice to show that if $\mu$ is invariant, so is its restriction $\mu|A$.
But it seems
$\int f \circ T d(\mu|A) = \int_A f \circ T d\mu$, and it seems $\int_A f \circ T d\mu$ may not equal $\int_A f d\mu$ despite $\int_M f d\mu = \int_M f \circ T d\mu$.
The question comes from this proof where they claim $\mu_0$ and $\mu_1$ are $\phi$-invariant.
Thanks!
Hint: Note that the subset you are conditioning on ought to be $T$-invariant as well; so that
$$\chi_{A}(x)\, f\circ T(x)=\chi_{T^{-1}(A)}(x)\, f\circ T(x) = \chi_A\circ T(x)\, f\circ T(x) = (\chi_A\, f)\circ T(x).$$
(Also the collection of measurable $T$-invariant subsets is a $\sigma$-algebra.)