Is Lebesgue integral a continuous functional in this context? Statement: I asked a similar question before which was closed. I tried to provide some context in this version.
Let $p$ be a probability measure whose support is $\left[a,b\right]$. Consider the set of continuous functions with a domain of $\left[a,b\right]$ and a codomain of $\left[c,d\right]$, denoted by $\mathcal{F}$. The reason I restrict the codomain is to ensure the dominated convergence theorem can be applied. Take $\int \cdot dp$ as a mapping from $\mathcal{F}$ to $\mathbb{R}$. If I endow $\mathcal{F}$ with the topology of pointwise convergence and $\mathbb{R}$ with the standard topology, is then $\int \cdot dp$ a continuous functional on $\mathcal{F}$?
Note that the integral and limit can be exchanged in this case. But since the topology of pointwise convergence is not metrizable, this is only a necessary condition for $\int \cdot dp$ to be continuous. I tried to go back to the definition of continuity for the general topological space, that is, the preimage of an open set is an open set. But that means when the value of the integral  is within a union of several open intervals, the value of the functions must be constrained in an open set at several points (I don't know how to describe this precisely but what I mean is the notion of subbasis for the topology of pointwise convergence on page 281 of Topology (2nd edition) by Munkres) and I have no idea how to verify that. I was wondering whether there are some ways to know whether $\int \cdot dp$ is a continuous functional or not.
 A: Consider $[0,1]$ with Lebesgue measure. Consider the net $(\chi_F)_{F \in \mathcal F}$  where $\mathcal F$ is the collection of all finite subsets of $[0,1]$ ordered by inclusion. This net converges to the constant function $1$ pointwise. $\int \chi_F =0$ for every $F$, but $\int 1 =1$. Hence, integral is not continuous for pointwise convergence topology.
[For pointwise convergence please refer to my comment below].
EDIT
OP wants continuous fuctions. This can be achieved as follows:
For any finite set $F \subseteq [0,1]$ there is a bounded continuous function $g_F$ which is $0$ at points of $F$ and strictly positive at all other points. By multiplying this function by a constant we can assume that the integral of this function is $1$. Let $f_F=1-g_F$. Then $(f_F)$ converges pointwise to $1$  and integral of $f_F$ is $0$ for every $F$.
Additional information: It can be shown that a linear map $L: C(X) \to \mathbb R$ is continuous for the topology of pointwise convergence if and only if there is a finite set $\{x_1,x_2,...,x_n\}\subseteq X$  and real numbers $c_1,c_2,...,c_n$ such that $L(f)=\sum\limits_{k=1}^{n} c_kf(x_k)$ for all $f \in C(X)$
