Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of a measures. We know, by the Portmanteau Theorem, that:
$$\int f d \mu_n \to \int f d\mu, \quad \forall \, f \in C_b \hbox{(class of continuous and bounded function)}$$ is equivalent to $\mu_n(E) \to \mu(E)$, for all $E \in \mathcal{C}_\mu$, class of continuity sets of $\mu$.
Now, I want show that
\begin{equation}\label{I}\tag{I}
\int f d \mu_n \to \int f d\mu, \quad \forall \, f \in C_{b}, \hbox{ vanishing on a neighborhood of } 0
\end{equation}
is equivalent to
\begin{equation}\label{II}\tag{II}
\mu_n(E) \to \mu(E), \quad E \in \mathcal{C}_{\mu}, \,\, 0 \notin \bar{E}
\end{equation}
where $\bar{E}$ denotes the closure of $E$.
Is this equivalence true?
Update
Let's try to give a stretch of proof of (\ref{II}) implies (\ref{I}). The converse was given in Jochen Wengenroth's answer.
Assume (\ref{II}) is valid and let any fixed $f \in \mathcal{C}_b$ vanishing on a neighborhood of $0$. Denote such neighborhood as $V_f$. Define for all $E \in \mathcal{C}_\mu$:
$$\nu(E) := \mu(E \cap V_f^c),\quad \nu_n(E) := \mu_n(E \cap V_f^c)$$
Since $0 \notin \overline{E \cap V_f^c}$ and $E \cap V_f^c \in \mathcal{C}_\mu$, using (\ref{II}), we have that $\nu_n(E) \to \nu(E)$, as $n \to \infty$ for all $E \in \mathcal C_\nu = \mathcal C_\mu$. By the Portmanteau Theorem, we have that
$$\int \bar f \chi_{V_f^c} d \mu_n = \int \bar f d \nu_n \to \int \bar f d \nu = \int \bar f \chi_{V_f^c} d \mu , \quad \bar f \in \mathcal C_b$$
So taking any $\bar f \in \mathcal C_b$ such that $\bar f|_{V_f^c} \equiv f|_{V_f^c}$, we have that $\bar f \chi_{V_f^c}= f$. This shows (\ref{I}) for $f$.
Best Answer
It seems that this can be deduced from the Portmanteau theorem: Assume the convergence of the intergals $\int fd\mu_n$ for all $f\in C_b$ vanishing in a neighbourhood of $0$ and fix $E\in C_\mu$ with $0\notin \overline E$. You may then choose a continuous function $g$ with values in $[0,1]$ which is $1$ in a neighbourhood of $E$ and $0$ in a neighbourhood of $0$. Then $E$ is also a continuity set of the measure $g\cdot \mu (A)=\int_Agd\mu$ and the measures $g\cdot\mu_n$ satisfy $\int fd(g\cdot \mu_n) = \int fgd\mu_n \to \int fgd\mu=\int fd(g\cdot \mu)$ for every $f\in C_b$ since $fg$ vanishes in a neighbourhood of $0$. Hence, $\mu_n(E)=(g\cdot \mu_n)(E)\to (g\cdot\mu)(E)=\mu(E)$.
The other implication is very similar.