Let $f:X\to X$ be a homemmorphim on a compact metric space $X$. Assume that $T:X \to \mathbb{R}$ is continuous. We consider $$ \sigma_{f}(x):=\lim_{n\to \infty}\frac{1}{n}\sum_{k=0}^{n-1}T(f^{k}(x).$$
$\textbf{Question}$: Why is $$W(\alpha):=\{x \in X, \sigma_{f}(x)=\alpha\}$$ not compact?
$W(\alpha)$ is the set of point where the limit of $\sigma_{f}(x)$ exists. I know the set might be an empty set(consider the identity functions), so consider the question in case the set is not empty.
I assume you mean "not always compact" (since for the identity map $\sigma_{Id}(x)=T(x)$ and so $W(\alpha)=T^{-1}(\alpha)$ which is a (possibly empty) compact set).
I don't know what kind of answer in terms of "why" you may be looking for, but one type of example is that $T$ has a repelling fixed point $x_0$ and all the nearby $x$ proceed to a different Birkhoff average, while $x_0$ stays and gets its own average. Consider:
Let $X=[0,1]$ and $f(x)=x^2$, and $T(x)=x$. We then have $\sigma_f(x)=\lim \frac{1}{n}(x+x^2+x^4+\ldots+x^{2^{n-1}})$ which is $1$ if $x=1$ and is zero for all other $x$. This means $W(0)=[0,1)$ is not compact.