I'll answer the topological version. Let's work in one dimension for simplicity, so we consider the space $\newcommand{\PR}{\mathcal{P}(\mathbb{R})}\PR$ of probability measures on $\mathbb{R}^1$.

Most of the commonly encountered metrics on $\PR$ induce the same topology: the weak topology. The standard reference for facts about this topology is Billingsley, *Convergence of Probability Measures*. The basic definition is that a sequence of measures $\mu_n$ converges weakly to a measure $\mu$ if $\int f\,d\mu_n \to \int f\,d\mu$ for all bounded continuous functions $f : \mathbb{R} \to \mathbb{R}$. However, for our purposes, a convenient fact is that when working on $\mathbb{R}$ (or any separable locally compact metric space), "bounded continuous" can be replaced with "continuous and compactly supported".

I claim the set of measures having bounded support is dense in $\PR$. (In particular all such measures have finite fourth moment.) To see this, fix a measure $\mu \in \PR$ and let $\mu_n(A) = \frac{\mu(A \cap [-n,n])}{\mu([-n,n])}$. Clearly $\mu_n$ is a probability measure with bounded support. Let $f \in C_c(\mathbb{R})$ and choose $N$ so large that $f$ is supported in $[-N,N]$. Then for all $n \ge N$ we have
$$\int f\,d\mu_n = \frac{1}{\mu([-n,n])} \int f\,d\mu \to \int f\,d\mu$$
as $n \to \infty$. So $\mu_n \to \mu$ weakly.

On the other hand, the space of probability measures $\newcommand{\PF}{\mathcal{P}^4(\mathbb{R})}\PF$ with finite fourth moment is meager in the weak topology. Let $A_k$ be the set of all probability measures having (uncentered) fourth moment $\le k$, so that $\PF = \bigcup_n A_k$. I claim $A_k$ is closed and nowhere dense.

Let $\mu_n \in A_k$ and suppose $\mu_n \to \mu$ weakly. Set $f_j(x) = \min(x^4, j)$ so that $f_j$ is bounded and continuous. We have $\int f_j\,d\mu_n \le \int x^4\,d\mu_n \le k$ for all $j,n$. Letting $n \to \infty$ we have $\int f_j\,d\mu = \lim_{n \to \infty} \int f_j\,d\mu_n \le k$. Now letting $j\to \infty$, we have $\int x^4\,d\mu = \lim_{j \to \infty} \int f_j\,d\mu$ by monotone convergence, so $\int x^4\,d\mu \le k$ and we have $\mu \in A_k$. Thus $A_k$ is closed.

Suppose $\mu \in A_k$ and let $\mu_n = (1 - \frac{1}{n}) \mu + \frac{1}{n} \delta_n$ where $\delta_n$ is a point mass at $n$. An argument like the above shows that $\mu_n \to \mu$ weakly, but you can also see that $\int x^4\,d\mu_n \ge n^3$, so for sufficiently large $n$ we have $\mu_n \notin A_k$. Thus $\mu$ is not in the interior of $A_k$; $\mu$ was arbitrary so $A_k$ is nowhere dense.