Given a sequence of independent but not necessarily identically distributed random variables $(X_n)$ with $\mathbb{E}X_n = 0$ for all $n$, the strong Law of Large Numbers says that $(1/n)\sum_{i=1}^n X_i \rightarrow 0$ almost surely.

Now let $S$ be a metric space and let $T$ be the shift operator on $S^\mathbb{N}$, that is $T(x_0,x_1,\dots) = (x_1, x_2,\dots)$. Consider a product measure $\mu = \prod_{i=0}^\infty \mu_i$ on $S^\mathbb{N}$ (thus $\mu$ is generally not invariant unless $\mu_i = \mu_0$ for all $i$).

For a given bounded, continuous function $f$ on $S^\mathbb{N}$ of the form $f = g \circ \pi$, where $g$ is a continuous bounded function on $S$ and $\pi(x_0,x_1,\dots) = x_0$, let $$X^f_n(\boldsymbol{x}) := f\circ T^n(\boldsymbol{x}) - \int f\circ T^n(\boldsymbol{y})\,\mu(\mathrm{d}\boldsymbol{y}).$$
On the probability space $(\Omega, \mathbb{P}) := (S^\mathbb{N}, \mu)$, we have that $(X_n^f)$ is a sequence of independent random variables with $\mathbb{E}X_n^f = 0$ for all $n$. Thus $(1/n)\sum_{i=0}^{n-1}X_n^f \rightarrow 0$ ($\mu$ a.e.).

Therefore, given a product probability measure on $S^\mathbb{N}$, a property analogous to ergodicity (w.r.t. $T$) holds when we restrict ourselves to continuous bounded maps of the form $f = g\circ\pi$ as above.

My question is: does this property still hold for the general $f\in C_b(S^\mathbb{N})$?