A function $f:\mathbb{R} \to \mathbb{R}$ is periodic if there exists some number $t > 0$ such that
$$ f(x) = f(x + t)
$$
A constant function is periodic since you can take $t = 1, t = 2$, etc. (Hint: Hover over the tag "periodic-functions". What do you see?)
The fundamental period of $f$ is the smallest of such $t$'s. Since $t$ cannot be $0$, you are looking for the minimum of $(0,\infty)$, which does not exist.
This should only be taken as a partial answer, since the results quoted are all "mod null", i.e. statements should only be understood to be true up to a set of measure zero. Probably a reader equipped with more descriptive set theory expertise would than me would know whether this can be upgraded to pointwise statements.
It is a result quoted, for example, in Brenier's "Polar Factorization and Monotone Rearrangement of Vector-Valued Functions", that if $(X,\mu)$ is a probability space (e.g a bounded domain in $\mathbb{R}^n$ with normalized Lebesgue measure), and $u\in L^p(X,\mu)$, then there exists a unique nondecreasing rearrangement $u^* \in L^p(0,1)$. Moreover, there exists a measure-preserving map $s$ from $(0,1)$ to $(X,\mu)$ such that $u\circ s = u^*$. In the case where X is understood to be a bounded subset of $\mathbb{R}^n$, this can be thought of as a rearrangement of the domain. (Brenier actually quotes the opposite result, but the paper he cites, by Ryff, gives both directions.)
By modifying $u^*$ on a null set, we can take $u^*$ to be lower-semicontinuous. In this case, it is apparent that $u^*$ is continuous iff it has no jump discontinuities.
In the case where $n>1$ one can also get some mileage out of Sobolev space theory. It follows from Theorem 0.1 of "A Co-area Formula with Applications to Monotone Rearrangement and to Regularity" by Rakotoson and Temam that if $u\in W^{1,p}(\Omega)$, then $u^* \in W^{1,p} (0,1)$. In particular, (up to choice of pointwise representative) $u^*$ is absolutely continuous, and therefore continuous. (This is not interesting in the case where $n=1$ since we would already be assuming that u is continuous.)
Best Answer
The usual definition of a periodic function presented in undergraduate calculus classes is something like
In particular, $f(x)$ must be defined for all real numbers. Hence every function that is periodic has, by definition, an unbounded domain. Implicit in the definition is that $f(x+kT) = f(x)$ for all integers $k$. As a pathological counterexample, consider the following pathological function:
For each $n\in\mathbb{N}$, choose (inductively) some $x_n \in (0,1)$ such that $x_i - x_j$ is irrational for all $i\ne j$. Define a function $f$ by setting $$ f(x_n + n + m) = 1, $$ where $m$ and $n$ range over the natural numbers. This function is pretty far from periodic, if you ask me (it doesn't really repeat itself in any meaningful way from one interval to the next), but if you drop the requirement that a periodic function be defined for all of $\mathbb{R}$, you could perhaps argue that $f$ is 1-periodic.
Another approach is to consider that a periodic function is really defined on an $n$-dimensional torus (a circle, in the case of functions that we want to think of as taking real inputs, but we can also talk about periodicity in $\mathbb{R}^n$). A periodic function on $\mathbb{R}$ then becomes some kind of lifting of a function on the circle via a covering map. Again, this would imply that the domain (thought of as a subset of $\mathbb{R}$) must be unbounded, though it needn't imply that the domain is all of $\mathbb{R}$.