Universal approximation of Gaussians Can gaussian kernels reproduce non continuous L2 integrable functions? ( Do non continuous L2 integrable functions lie in the RKHS constructed by a Gaussian Kernel?)
Edit:
I think my question is being misconceived. I do not intend to say that the RKHS is constructed using this argument. My question is that due to the reproducing property, every function that belongs to a Hilbert space must satisfy the reproducing property give the kernel is a reproducing kernel. Would a constant function over a compact support satisfy this for gaussian kernels?
 A: $L_2$ integrable functions are equivalence classes of functions that can differ on subsets of measure zero. For an RKHS, you need the evaluation function at a given point $x$ to be well defined, which is not possible for such an equivalence class, since a point is of measure zero, and thus all members of the equivalence class can have different values at $x$.

Edit:
When I read in the question, that the OP was concerned with noncontinuous functions in a vector space complete w.r.t. the $L_2$ norm, my immediate reflex was to think of a space of equivalence classes of functions. And in such a space, evaluation at a point is in general not well defined, hence my answer.
But, as @whuber pointed out in the comments, there are $L_2$ Hilbert spaces that don't need equivalence classes to be defined properly and for which evaluation at a point is well defined. I can also think of those that, as requested by the OP, contain noncontinuous functions.
But the OP asked moreover for a Gaussian RKHS, i.e. one with a Gaussian kernel. And I was not able to think of a Gaussian L2 RKHS with noncontinuous functions.
So I consulted the paper:

Cucker, Felipe, and Steve Smale. "On the mathematical foundations of learning." Bulletin of the American mathematical society 39.1 (2002): 1-49.

There, in Chapter III, Section 3, Theorem 2 is relevant to the OP's question: It essentially states that, if we have a Mercer kernel, i.e. the kernel is continuous, symmetric, and positive definite, which is the case for the Gaussian kernel, then the RKHS consists of continuous functions.
Thus, to summarize, while my first reasoning was wrong, the statement stays correct: There are no noncontinuous functions in the $L_2$ RKHS of a Gaussian kernel.
A: More generally, a Hilbert space $\mathcal F$ of functions from $\mathcal X$ to $\mathbb R$ is a Reproducing Kernel Hilbert Space if and only if for all $x\in\mathcal X$, the map :
$$L_x : f\in\mathcal F\mapsto f(x)$$
is bounded. That is, if and only if there exists for all $x\in\mathcal X$ a finite $M_x$ such that
$$|f(x)|\le M_x \|f\|_{\mathcal F}\quad \forall f\in\mathcal F \tag1$$
So even if you forget for a minute that for $\mathcal F=\mathcal L^2(\mathcal X)$, $f(x)$ is not well-defined (as rightfully mentioned in @frank's answer), you can easily see that the uniform bound $(1)$ doesn't hold. For instance, let $x_0\in\mathcal X$ be arbitrary and consider the family of functions
$$f_n(x) =\begin{cases} n \text{ if } x=x_0,\\
 0 \text{ otherwise}\end{cases}$$
It is clear that all the $f_n$'s are in $\mathcal F =\mathcal L^2$ but the evaluation functional $L_{x_0}$ is unbounded by construction. Hence we can conclude that $\mathcal L^2$ is not a RKHS for any kernel, and in particular for the Gaussian kernel.
By a similar argument you can show that even a "well-behaved" space such as the set of all continuous functions on an interval $[a,b]$ is not an RKHS either.
