TITLE: Open set in such a set w.r.t weak topology
QUESTION [7 upvotes]: Let $(\mathcal X,\mathcal F)$ be a measurable space and $\mathcal P$ the set of probability measures  on this space. With weak convergence (i.e., $P_n\to P$ weakly if and only if $\mathbb E_{P_n}(f)\to\mathbb E_P(f)$ for every bounded continuous functions $f:\mathcal X\to\mathbb R$), a basis of the corresponding weak topology induced by this weak convergence is a collection of the following sets:
$$U(P,\alpha)=\left\{Q:\left|\int f_idP-\int f_idQ\right|<\alpha,i=1,\ldots,n\right\}$$
for some $P\in\mathcal P$, $f_i$ bounded continuous functions, $n$ is finite, and $\alpha>0$. Then there exists a finite measurable partition $A_1,\ldots,A_m$ of $\mathcal X$ and $\beta>0$ such that
\begin{align}
\tilde{U}(P,\beta)=\{Q:|P(A_j)-Q(A_j)|<\beta\}\subset{U}(P,\alpha).
\end{align}
My questions are:

is $\tilde{U}(P,\epsilon)$ also open? 
if not, is there an open set, say $U'(P, \alpha')$, that is contained in $\tilde{U}(P,\beta)$? or under what conditions there exists such a $U'(P, \alpha')$? (might have different bounded continuous functions)

Can you help me with these question? or please see my thinking below.

Let me first prove the existence of such $\tilde{U}(P,\beta)\subset U(P,\alpha).$
Since $f_i$ is bounded continuous, then there exists a simple function $\tilde f_i$ such that $\|f_i-\tilde f_i\|_{\infty} <\frac{\alpha}{3}$. Then if $|\int \tilde{f}_idP - \int \tilde{f}_idQ|<\frac{\alpha}{3}$, we have 
\begin{align}
&\left|\int f_idP-\int f_idQ\right|\\
=~&\left|\int f_idP-\int \tilde f_idP+\int \tilde f_idP-\int \tilde f_idQ+\int \tilde f_idQ-\int f_idQ\right|\\
<~&\alpha.
\end{align}
We can then define a finite partition according to these simple functions, as $f_i$'s are bounded. By choosing a small enough $\beta$, $|P(A_j)-Q(A_j)|<\beta$ for all $i$ implies $|\int \tilde f_idP-\int \tilde f_idQ|<\frac{\alpha}{3}$.

I have no idea about question 1, so I go to question 2, thinking to construct some bounded continuous functions to build a new open set.
a). for each measurable set $A_i$, find two continuous functions $\bar{f_i}$ and $\underline{f_i}$, such that 
$$\bar{f_i}(x)\geq I_{A_i}(x)\geq \underline{f_i}(x),~\text{for every $x\in\mathcal X$},$$
$$\int \bar{f_i} dP- \int \underline{f_i}dP<\frac{\beta}{2},$$
where $I_{A_i}(\cdot)$ is the indicator function.
b). For any probability measure $Q$, we have $$\int\underline{f_i}dQ\leq Q(A_i)\leq\int\bar{f_i}dQ.$$ Thus, if $|\int\bar{f_i}dP - \int\underline{f_i}dQ|<\beta$ and $|\int\underline{f_i}dP - \int\bar{f_i}dQ|<\beta$, we must have $|P(A_i)-Q(A_i)|<\beta$. We also have
\begin{align}
&\left|\int\underline{f_i}dP - \int\bar{f_i}dQ\right|\\
=~&\left|\int\underline{f_i}dP-\int\bar{f_i}dP+\int\bar{f_i}dP-\int\bar{f_i}dQ \right|\\
\leq~&\left|\int\underline{f_i}dP-\int\bar{f_i}dP\right|+\left|\int\bar{f_i}dP-\int\bar{f_i}dQ \right|
\end{align}
Similarly, we have $$\left|\int\bar{f_i}dP - \int\underline{f_i}dQ\right|\leq\left|\int\bar{f_i}dP-\int\underline{f_i}dP\right|+\left|\int\underline{f_i}dP-\underline{f_i}dQ \right|$$
As such, adding the following constraints $$\left|\int\underline{f_i}dP-\underline{f_i}dQ \right|<\frac{\beta}{2}, \left|\int\bar{f_i}dP-\bar{f_i}dQ \right|<\frac{\beta}{2}$$
will construct a new open set that is contained in $\tilde{U}(P,\beta)$.
It seems that a) is not always true. When $\mathcal X=\mathbb R$, consider $P= \frac{1}{2}I_{[0,1]} + \frac{1}{2}I_{\{3\}}$. For a measurable set $A=[2,3)$, no such $\bar{f_i}$ and $\underline{f_i}$ exist for each $\beta>0$. I think it possible to place some constraints on the partition $A_1,A_2,\ldots, A_m$ for the simple functions $\tilde f_i$, but I got no formal proof/reference to it. Can you advise me how to make this work? Thanks!!~

REPLY [2 votes]: This is not true, let $X=[-1,1]$ and consider $P = \delta_0$. Let $f_1,...,f_n$ be continuous on $X$. Note that $U_{\{f_1,..,f_n\}}(P,\epsilon)=\bigcap_i U_{f_i}(P,\epsilon)$.
You can do a continuity argument to find non-singular measures $Q=\frac1h \chi_{[-h,h]}$ so that for any fixed $f$ you have $Q\in U_f(P,\epsilon)$. So any weak-open set around $f$ contains non-singular measures. On the other hand let $A=\{0\}$, then
$$\tilde U_A (P,\epsilon) = \{Q\mid |1-Q(\{0\})|<\epsilon\}$$
contains only measures that are singular (provided $\epsilon<1$), thus this cannot be an open neighbourhood of $P$.
If you restrict your $A_i$ to be open the statement should be true (if your measure space has the Borel sigma algebra), if necessary add in $X$ to be locally compact and probability measures to be regular.