No. For example, take $S=\mathbb R$ and let $\mu_n$ be Lebesgue measure restricted to $[n,n+1]$. Set $$ f(x_0,x_1,\ldots)=\sin^2(2\pi x_0)\sin(2\pi(x_{2^j-\lfloor x_0\rfloor}) \text{ if $x_0\in [2^{j-1},2^j)$}. $$

Then $f(T^n(x_0,x_1,\ldots))=\sin^2(2\pi x_0)\sin(2\pi x_{2^j})$ for all $n$ in the range $[2^{j-1},2^j)$ (the reason for the $\sin^2$ term is just to make it continuous). The averages don't converge.

No. For example, take $S=\mathbb R$ and let $\mu_n$ be Lebesgue measure restricted to $[n,n+1]$. Set $$ f(x_0,x_1,\ldots)=\sin^2(2\pi x_0)\sin(2\pi(x_{2^j-\lfloor x_0\rfloor}) \text{ if $x_0\in [2^{j-1},2^j)$}. $$

Then $f(T^n(x_0,x_1,\ldots))=\sin^2(2\pi x_n)\sin(2\pi x_{2^j})$ for all $n$ in the range $[2^{j-1},2^j)$ (the reason for the $\sin^2$ term is just to make it continuous). The averages don't converge.

No. For example, take $S=\mathbb R$ and let $\mu_n$ be Lebesgue measure restricted to $[n,n+1]$. Set $$ f(x_0,x_1,\ldots)=x_{2^j-\lfloor x_0\rfloor}-2^j \text{ if $x_0\in [2^{j-1},2^j)$}. $$

Then $f(T^n(x_0,x_1,\ldots))=x_{2^j}-2^j-\frac 12$ for all $n$ in the range $[2^{j-1},2^j)$. The averages don't converge.

No. For example, take $S=\mathbb R$ and let $\mu_n$ be Lebesgue measure restricted to $[n,n+1]$. Set $$ f(x_0,x_1,\ldots)=\sin^2(2\pi x_0)\sin(2\pi(x_{2^j-\lfloor x_0\rfloor}) \text{ if $x_0\in [2^{j-1},2^j)$}. $$

Then $f(T^n(x_0,x_1,\ldots))=\sin^2(2\pi x_0)\sin(2\pi x_{2^j})$ for all $n$ in the range $[2^{j-1},2^j)$ (the reason for the $\sin^2$ term is just to make it continuous). The averages don't converge.