WARNING: The following post certainly contains too much information. I've included the complete context of my question so that it is clear what I'm talking about and where my question comes from. Feel free to jump to the last two block quotes (that is probably enough).
I am reading "Introduction to Ergodic Theory", by Y. Sinai. The book is short on details and often contains statements that leave me wondering why they are true.
On page 12, he writes
Let $X$ be a compact topological space and let $T$ be a continuous transformation of $X$ into itself. Then there always exists at least one measure on $X$ that is invariant with respect to $T$.
Proof: Let $\nu$ be an arbitrary normalized measure.
Here I assume he is taking the $\sigma$-algebra $\mathcal{B}$ of measurable subsets of $X$ to be the Borel $\sigma$-algebra (the smallest $\sigma$-algebra containing the open sets), and $\nu$ is an arbitrary measure such that $\nu(X)=1$.
(continued) We set $\nu_k(A)=\nu(T^{-k}A)$, or $T^k\nu=\nu_k$. Furthermore, we introduce the measure $\overline{\nu}_n=\frac{1}{n}\sum_{k=0}^{n-1}\nu_k$.
As I understand it, all of the measure he has just introduced are defined on the same Borel $\sigma$-algebra $\mathcal{B}$, and they are all normalized (i.e. probability) measures.
(continued) The space of normalized measures on a compact space is weakly compact.
I understand this to mean that the set of measures $\{\nu:\mathcal{B}\rightarrow[0,\infty]\;|\;\nu(X)=1\}$ is compact in the weak* topology (that is, the smallest topology such that the function $\nu\mapsto\int_Xf\;d\nu$ is continuous for all measurable $f:X\rightarrow\mathbb{R}$). And I think this is clear since now $\nu\mapsto\int_X1\;d\nu$ is continuous, and its inverse image is precisely the set of normalized (i.e. probablity) measures on $X$.
(continued) This means that a normalized measure $\mu$ and a strictly monotone sequence of numbers $n_i$ can be found such that $\lim_{n_i\rightarrow\infty}\nu_{n_i}=\mu$.
OK, this is sequential compactness.
(continued) We will prove the invariance of the measure $\mu$, i.e. the equality $\int f(x)\;d\mu=\int f(Tx)\;d\mu$ for any bounded measurable function $f$.
This is fine since the author has previously shown that invariance of measure ($\mu(A)=\mu(T^{-1}A)$ for all measurable $A$) is equivalent to equality of the above integrals.
(continued) By virtue of the compactness of the space $X$ it is sufficient to verify the last relation only for continuous functions (they form an everywhere dense set in the space of all measurable functions).
AND HERE IS MY QUESTION: Why is this last sentence true? I am asking both: (1) why is the statement in parenthesis true? and (2) how would one show that $\int f(x)\;d\mu=\int f(Tx)\;d\mu$ for any bounded measurable function $f$ if one knew already that $\int f(x)\;d\mu=\int f(Tx)\;d\mu$ for any continuous function $f$?
Thank you.