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 $f$.



Show that a certain convergence of measures is equivalent to a certain convergence of integrals
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.


