Prove that $d_k=\sup_{f \in C_{\leq 1}^k} \bigg\vert \int_\mathbb{R} f \ d \mu - \int_\mathbb{R} f \ d \nu \bigg\vert$ satisfies Hausdorff property Let $ C_b^k(\mathbb{R},\mathbb{R}) $ denote the bounded, continuous functions from $ \mathbb{R} $ to $ \mathbb{R}$ which are $ k \in \mathbb{N}_0 $ times differentiable with bounded derivatives. Define
$$
C_{\leq 1}^k \equiv \bigg\{ f \in C_b^k(\mathbb{R},\mathbb{R}) \colon \sum_{i=0}^k \Vert f^{(i)} \Vert_\infty \leq 1 \bigg\},
$$
where $\Vert \cdot \Vert_\infty$ is the sup norm of $ f $ and define
$$
d_k(\mu, \nu) \equiv \sup_{f \in C_{\leq 1}^k} \bigg\vert \int_\mathbb{R} f \ d \mu - \int_\mathbb{R} f \ d \nu \bigg\vert
$$
where $ \mu, \nu $ are probability measures on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$.

I want to prove that $d_k$ is a metric. We note that  $ d_k $ is symmetric and non-negative since $ (x,y) \mapsto \vert x - y \vert $ is symmetric and non-negative. Also we see that $d_k$ satisfies the triangle inequality since for probability measures $\mu,\nu,\rho$,  we have
\begin{align*}
d_k(\mu,\nu) &= \sup_{f \in C_{\leq 1}^k} \bigg\vert \int_\mathbb{R} f \ d \mu - \int_\mathbb{R} f \ d \nu \bigg\vert \\[1em]
&=  \sup_{f \in C_{\leq 1}^k} \bigg\vert \int_\mathbb{R} f \ d \mu - \int_\mathbb{R} f \ d \rho + \int_\mathbb{R} f \ d \rho - \int_\mathbb{R} f \ d \nu \bigg\vert \\[1em]
& \leq \sup_{f \in C_{\leq 1}^k} \bigg( \bigg\vert \int_\mathbb{R} f \ d \mu - \int_\mathbb{R} f \ d \rho \bigg\vert + \bigg\vert\int_\mathbb{R} f \ d \rho - \int_\mathbb{R} f \ d \nu \bigg\vert \bigg) \\[1em]
& \leq \sup_{f \in C_{\leq 1}^k}  \bigg\vert \int_\mathbb{R} f \ d \mu - \int_\mathbb{R} f \ d \rho \bigg\vert + \sup_{f \in C_{\leq 1}^k} \bigg\vert\int_\mathbb{R} f \ d \rho - \int_\mathbb{R} f \ d \nu \bigg\vert \\[1em]
&= d_k(\mu,\rho) + d_k(\rho,\nu).
\end{align*}

However, I have a lot of trouble proving the Hausdorff propterty i.e.
$$
d_k(\mu,\nu) = 0 \iff \mu = \nu.
$$
Of course "$\Leftarrow$" is easy but it is "$\Rightarrow$" that causes me trouble. Assuming that $d_k(\mu,\nu) = 0$ we get that
$$
\int_\mathbb{R} f \ d \mu = \int_\mathbb{R} f \ d \nu
$$
for all $f \in C_{\leq 1}^k$. I have a result stating that if for all $f \in C_b(\mathbb{R})^{+}$ we have
$$
\int_\mathbb{R} f \ d \mu = \int_\mathbb{R} f \ d \nu
$$
then $\mu=\nu$. However how can I be sure that $C_{\leq 1}^k $ is a "broad" enough class of functions to conclude as well that $\mu=\nu$? Can I prove this using the known result I have stated? Or is there a better way to go about it?
 A: Note, by scaling, that you really have $\int_{\mathbb{R}} f\textrm{d}\mu=\int_{\mathbb{R}} f\textrm{d}\nu$ for all $f\in C^k_b$, that is, the $C^k$ functions with $k$ bounded derivatives.
Hence, all you need is to know that this class is dense in $L^1(\mu)$ and in $L^1(\nu)$. There are many ways of proving it. My prefered way is showing that they can approximate indicators of the form $1_{[a,b]}$.
Indeed, let $\phi(t)=1_{t>0}(t) e^{-1/t}$. You may check that $\phi$ defines a smooth function. Furthermore,
$$
\psi(t)=\frac{\phi(t)}{\phi(t)+\phi(1-t)}
$$
also defines a smooth function (its denominator is never $0$), $\psi|_{(-\infty,0])}\equiv 0$ and $\psi|_{[1,\infty)}\equiv 1$.
This allows us to define $$
\xi_{\varepsilon}(t)=\begin{cases} \psi(\frac{t}{\varepsilon}) & t\leq \varepsilon \\ 1 & t\in (\varepsilon,1+\varepsilon) \\ \psi(2+\frac{1-t}{\varepsilon}) & t\geq 1+\varepsilon\end{cases}
$$
which, again is a smooth function for every $\varepsilon>0$ and $\xi_{\varepsilon}\to 1_{[0,1]}$ pointwise. Note, furthermore, that since $\xi_{\varepsilon}$ is constant outside of a compact interval, it has bounded derivatives of all orders.
Now, since $|\xi_{\varepsilon}|\leq 1$, the dominated convergence theorem implies that
$$
\mu([0,1])=\int_{\mathbb{R}} 1_{[0,1]}\textrm{d}\mu=\lim_{\varepsilon\to 0^+} \int_{\mathbb{R}} \xi_{\varepsilon}\textrm{d}\mu=\lim_{\varepsilon\to 0^+} \int_{\mathbb{R}} \xi_{\varepsilon}\textrm{d}\nu=\int_{\mathbb{R}} 1_{[0,1]}\textrm{d}\nu=\nu([0,1])
$$
It's easy to use $\xi_{\varepsilon}$ to construct an approximating sequence for other values of $a$ and $b$ and thus, you get your result by the uniqueness theorem for probability measures.
