Let M be a metric space, $f: M \rightarrow M$ be a measurable transformation and $\mu$ be a measure on M. Show that $f$ preserves $\mu$ if and only if $\int \phi d\mu =\int \phi \circ f d\mu$ for every bounded continuous function $\phi: M \rightarrow \mathbb{R}$.
My attempt at a proof: First prove this direction $\rightarrow$.
We assume that $f \mbox{ preserves } \mu$
I have proved before that If $f:M \rightarrow M$ is a measurable transformation and $\mu$ is a measure on $M$. Then $f$ preserves $\mu$ if and only if $\int \phi d\mu = \int \phi \circ f d\mu$ for every integrable $\phi: M \rightarrow \mathbb{R}$. (This was easy to prove).
Thus since bounded continuous functions are dense in $L^1$, (i.e. $C(M) \subset L^1(M,\mu))$.
Now for the other direction. Suppose that $\int\phi d\mu=\int \phi \circ f d\mu$ for every bounded continuous function $\phi: M \rightarrow \mathbb{R}$. Need to show that $f$ preserves $\mu$, meaning that we need to show that $\mu (E) =\mu(f^{-1}(E))$ for every measurable set $E \subset M$.
Now, I'm not sure how to show how this, relation holds $\mu (E) =\mu(f^{-1}(E))$ for every measurable set $E \subset M$, where f is a continuous bounded function. I know how to do it when $f$ is just a characteristic function (then simple etc..). But, I don't know how to show it is true for arbitrary f that is continuous and bounded. Do I need to use something with linear functionals or is there a way to do it without that? Thank you very much.
Your argument for the first part is not valid. How do you know that bounded continuous functions are dense in $L^{1}(\mu)$ for any Borel measure $\mu$ on any metric space?
You have assume that $\mu$ is a finite measure. Otherwise bounded continuous functions may not be integrable.
Here is how you can prove this result: let $\nu(E)=\mu (f^{-1}(E))$. This defines another finite measure and what we are asked to prove is that $\mu=\nu$ iff $\int \phi\, d\mu =\int \phi\, d\nu$ for all bounded continuous functions $\phi$. Suppose this equation holds for such $\phi$ and let $C$ be a closed set. For each $n$ let $A_n=\{x\in M: d(x,C) <\frac 1 n\}$; there exists a continuous function $f_n:M\to[0,1]$ such that $f_n=1$ on $C$ and $f_n(x) =0$ if $x \notin A_n$. Apply the hypothesis with $\phi$ replaced by $f_n$ and let $ n \to \infty$. Note that $f_n \to I_C$ pointwise. Use DCT to conclude that $\mu (C)=\nu (C)$. Thus $\mu$ and $\nu $ coincide on the class of all closed sets. Can you show that $ \mu=\nu$? [ One way is to use the $\pi -\lambda $ theorem].