Are the "bounded, uniformly continuous functions" analogous to test functions as seen in Schwartz Distributions? I am reading this passage from Billingsley's Convergence of Probability measures.

Theorem 1.2. Probability measures $P$ and $Q$ on $\mathcal{S}$ coincide if $P f=Q f$ for all bounded, uniformly continuous real functions $f$.
Proof. For the bounded, uniformly continuous $f$ of (1.1), $P F \leq$ $P f=Q f \leq Q F^\epsilon$. Letting $\epsilon \downarrow 0$ gives $P F \leq Q F$, provided $F$ is closed. By symmetry and Theorem $1.1, P=Q$.
Because of theorems like this, it is possible to work with measures $P A$ or with integrals $P f$, whichever is simpler or more natural. We defined weak convergence in terms of the convergence of integrals of functions, and in the next section we characterize it in terms of the convergence of measures of sets.

Are the "bounded, uniformly continuous" analagous to the "test functions" $C_c^\infty(X)$ seen in the "theory of distributions? Is this describing that we can think of probability measures as "distributions" or "measures" by the reisz representation theorem? Why do they use "bounded, uniformly continuous" real functions instead of infinitely differentiable continuous functions with compact support as used in Schwartz distributions?
https://en.wikipedia.org/wiki/Distribution_(mathematics)
 A: The starting point is: Two probability measures $P$ and $Q$ on the same measurable space $S$ coincide if for all measurable functions $f:S\rightarrow\{0,1\}$ we have $Pf=Qf$.
To see this, take an event $E\subseteq S$ and let $f$ be given by $f^{-1}(1)=E$. Then we have $P(E)=Pf=Qf=Q(E)$.
This can be weakened into several directions.
On the other hand, we may want to approximate these indicators with other functions, which may be convenient in certain situations.
Here, we choose to look at metric spaces $S$ and continuous functions, simply because this is very frequent, though still general. Notice that in this generality there exists no notion of derivative.
For the metric space we choose the consistent Borel algebra.
Now, we need to observe that we can indeed ensure that we obtain the result for the indicators from the result for continuous functions. So, for $E$ we consider the $\varepsilon$-neighborhood $E_\varepsilon$ (cf. Wikipedia). The arguably easiest approximation is $f_\varepsilon(s)=1-\frac{1}{\varepsilon}\min(\varepsilon,d(s,E))$. Notice that $Pf_\varepsilon-P(E_\varepsilon\setminus E)\le Pf\le Pf_\varepsilon$, which shows that $Pf=\lim_{\varepsilon\rightarrow 0}Pf_\varepsilon$ using continuity from above whenever $E$ is closed, thus also for open $E$ and further for all $E$ (cf. this answer).
Combining this with the first part gives that $P$ and $Q$ are equal if $Pf_\varepsilon=Qf_\varepsilon$ for all $E$ and all $\varepsilon$. Now, we can weaken this set to our liking. Reasonably weakening this gives equality if $Pf=Qf$ holds for all Lipschitz continuous $f:S\rightarrow[0,1]$. This can be further weakened to the statement in the question.
Notice that we can also smoothen the edges of the $f_\varepsilon$ above, in the sense that we smoothen the piecewise linear functions $d\mapsto 1-\frac{1}{\varepsilon}\min(\varepsilon,d)$, if this is convenient in a given context.
Summary: The measures are actually represented by the expectations of the indicators at the beginning, which is trivial. The class of functions is not of any particular importance in this context. In my experience, a very prominent class are bounded Lipschitz continuous functions (e.g. here) and I dare to claim that the result was stated for uniformly continuous functions mostly because Lipschitz continuous functions have not been introduced.
A: Let $\mu_n, \mu$ be Borel probability measures on a metric space $X$. We have the following equivalence (known as Portmanteau's theorem):

*

*$\int f d \mu_n \to \int f d \mu$ for all $f$ bounded continuous.

*$\int f d \mu_n \to \int f d \mu$ for all $f$ bounded uniformly continuous.

*$\int f d \mu_n \to \int f d \mu$ for all $f$ bounded Lipschitz continuous.

If $X$ is locally compact and separable, then $\int f d \mu_n \to \int f d \mu$ for all $f$ bounded continuous IFF $\int f d \mu_n \to \int f d \mu$ for all $f$ continuous with compact support.
If $X = \mathbb R^n$, then $\int f d \mu_n \to \int f d \mu$ for all $f$ bounded continuous IFF $\int f d \mu_n \to \int f d \mu$ for all smooth $f$ with compact support.

So for $X=\mathbb R^n$, what you hope for is indeed true.
