Birkhoff spectrum is not compact 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.
 A: Perhaps you need to realize that $\sigma_f$ need not be continuous under your assumptions.

Actually, in many situations it turns out that $W(\alpha)$ is dense but not compact for $\alpha$ in some interval. This is the case for example for the dynamics on a hyperbolic set and Hölder continuous functions.
The main reference for this type of results is L. Barreira and J. Schmeling, Sets of “non-typical” points have full topological entropy and full Hausdorff dimension, Israel J. Math. 116 (2000), 29–70.

Incidentally, you need of course to clarify what you mean by "the limit exists for almost every point".
A: 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.
