I am trying to relate weak convergence of a sequence of measures on $\mathbb R$ and the uniform convergence on compact subsets of their (almost) Fourier transform. Here's what I know :
Each measure $\mu_m$ happens to be the difference of 2 probability measures on $\mathbb R$.
Each function $\phi_m(x) = \int_{\mathbb R} {\mathrm e}^{2i\pi xy}{\mathrm d}\mu_m(y)$ defined on $\mathbb R$ is continuous, bounded and $\|\phi_m\|_{\infty} = 1$. The sequence $\left(\phi_m \right)_{\mathbb N}$ converges uniformly on compact subset to the $0$ function.
I want to prove that $\left(\mu_m \right)_{\mathbb N} \overset{weak} {\rightarrow} 0$.
Since $\left( \mu_m \right)_{\mathbb N}$ lives in a bounded set (the ball of center $0$ and radius $2$) of the measure algebra $\mathcal M(\mathbb R)$ (for total variation), we can test weak convergence against functions in $C_0\left(\mathbb R\right)$ rather than $C_b\left(\mathbb R\right)$.
Let $\epsilon > 0, \ f \in C_0\left(\mathbb R\right)$. My plan was to approximate $f$ in a good way by a function $h \in \mathcal S (\mathbb R)$ (Schwartz space) so that I can use Fubini and the inverse Fourier transform. I would write $$\int_{\mathbb R} f \ \mathrm d\mu_m = \int_{\mathbb R} h \ \mathrm d\mu_m + \int_{\mathbb R} \left(f-h \right) \mathrm d\mu_m$$ and $$\int_{\mathbb R} h(x) \ \mathrm d\mu_m(x) = \int_{\mathbb R} \int_{\mathbb R} \hat h (y) {\mathrm e}^{2i\pi xy} \mathrm d\mu_m(x) = \int_{\mathbb R} \hat h \phi_m \mathrm d \lambda$$
Now we know that $\hat h \in C_0\left(\mathbb R\right)$, so for a given $\epsilon$, the set $K_{\epsilon}= \{z \in \mathbb R \ |\ \left| \hat h (z)\right| \geq \epsilon \}$ is a compact subset of $\mathbb R$. So $$\int_{\mathbb R} \hat h \phi_m \mathrm d \lambda = \int_{\mathbb K_{\epsilon}} \hat h \phi_m \mathrm d \lambda + \int_{\mathbb R \setminus K_{\epsilon}} \hat h \phi_m \mathrm d \lambda$$
On $\mathbb R \setminus K_{\epsilon},\ \hat h$ is bounded by $\epsilon$ so by Hölder inequality, $$\int_{\mathbb R \setminus K_{\epsilon}} \hat h \phi_m \mathrm d \lambda = \int_{\mathbb R} \hat h \phi_m \mathbb{1}_{\mathbb R \setminus K_{\epsilon}} \mathrm d \lambda \leq \|\hat h \mathbb{1}_{\mathbb R \setminus K_{\epsilon}} \|_{1}\ \|\phi_m\mathbb{1}_{\mathbb R \setminus K_{\epsilon}} \|_{\infty} \leq \epsilon /4$$
On $K_\epsilon$, which is compact, $\phi_m$ converges uniformly to $0$, so there exists a $m_0 \in \mathbb N$ for which $\int_{K_{\epsilon}} \hat h \phi_m \mathrm d \lambda \leq \epsilon /4$
Thanks to Possible density of Schwartz Functions in the space of continuous functions vanishing at infinity , I expect to simply chose $h \in \mathcal S (\mathbb R)$ close enough to $f$ in $\|.\|_{\infty}$ distance so that $$\int_{\mathbb R} \left(f-h\right) \mathrm d \mu_{m_0} \leq \epsilon /2$$ by dominated convergence theorem with the finite measure $\mu_{m_0}$.
Is this last reasoning good ? Is the proof good ? Thank you in advance for your answers.