A point $x \in X$ is $\mu$ - generic if $$\lim \frac{1}{n} \sum_{j=0}^{n-1} \delta_{T^{n}x} = \mu$$ (limit in the weak sense).
Take, for example, the following construction:
$X = \{0,1\}^{\mathbb{Z}}$ equipped with product topology and shift map $(\sigma(x))_{n\geqslant 1} = x_{n+1}$;
my question is, is there a point $x$ not generic to any $\sigma$ - invariant measure?
My intuition says yes because you should be able to construct a sequence of $0$s and $1$s satisfying the invariant part while also choosing them so there is no convergence in the definition of a generic point. I just don't know how :-(
Take the following element of $\{0,1\}^\mathbb{N}$: $$ y=0^1 1^3 0^9 1^{27}...,$$ that is sequences of length $3^k$ of $0$ and $1$ alternatively. It's not generic for any $\mu$.
By contradiction, assume there is some shift-invariant probability measure $\mu$ for which $y$ is generic. Put $f(x)=x_0$, that is $0$ is the sequence begins with a zero, $1$ otherwise (it is continuous). Then we would have $$\mu(f)=\lim_{N\to \infty} \frac1N \sum_{k=0}^{N-1}f(T^k y),$$ but there is no limit to the proportion of 0's and 1's in $y$.
For example, if $N=\sum_{k=0}^{2p} 3^{k}$, the ratio is $$\frac1N \sum_{k=0}^{N-1}f(T^k y)=\frac{3\sum_{i=0}^{p} 9^i}{\sum_{k=0}^p 3^{k}}= 3\frac{9^{p}-1}{8}\frac{2}{3^{2p+1}-1} \to \frac14,$$ and for $N=\sum_{k=0}^{2p+1} 3^{k}$, the ratio is $$\frac1N \sum_{k=0}^{N-1}f(T^k y)=\frac{3\sum_{i=0}^{p} 9^i}{\sum_{k=0}^{2p+1} 3^{k}}=3\frac{9^{p+1}-1}{8}\frac{2}{3^{2p+2}-1}= \frac34,$$