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

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

$$$$\label{I}\tag{I} \int f d \mu_n \to \int f d\mu, \quad \forall \, f \in C_{b}, \hbox{ vanishing on a neighborhood of } 0$$$$

is equivalent to

$$$$\label{II}\tag{II} \mu_n(E) \to \mu(E), \quad E \in \mathcal{C}_{\mu}, \,\, 0 \notin \bar{E}$$$$ where $$\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 $$f$$.

• Do you assume probabibilty measures? Nov 13, 2022 at 10:51
• Not necessarily.
– PSE
Nov 14, 2022 at 2:38

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, $$\mu_n(E)=(g\cdot \mu_n)(E)\to (g\cdot\mu)(E)=\mu(E)$$.