# Largest weak(-like) topology with respect to which continuous functions are dense in the space of Borel functions

Let $$X$$ denote the space of bounded Borel functions $$f\colon [0,1] \to \mathbb{R}$$. Let $$M$$ denote the space of finite Borel measures on $$[0,1]$$. What is the largest family $$F \subset M$$ such that for any $$f\in X$$ there exists a sequence $$(f_n)_{n=1}^\infty \subset C[0,1]$$ such that for any $$\mu \in F$$ $$\int_{0}^{1} f_n(t) \, d\mu(t) \to \int_{0}^{1} f(t) \, d\mu(t) \quad as \quad n\to \infty,$$ i.e. $$f_n \to f$$ in the topology $$\sigma(X,F)$$.

In other words, what is the largest family $$F \subset M$$ that $$C[0,1]$$ is sequentially dense in $$(X, \sigma(X, F)))$$?

Clearly $$F$$ includes all absolutely continuous (with respect to Lebesgue) measures, because any bounded Borel function can be approximated with uniformly bounded continuous functions converging strongly in $$L^1$$. On the other hand $$F\not = M$$ as pointed out in a related post (which is actually a motivation for this question).

Update. If we asked $$C[0,1]$$ be dense (not sequentially dense) in $$(X, \sigma(X,F))$$, then one could have taken simply $$F=M$$. Indeed, for any finite set of measures $$\mu_1$$, ..., $$\mu_n$$ from $$M$$ we know that $$C[0,1]$$ is dense in $$L^1([0,1], |\mu_1| + \ldots + |\mu_n|)$$. But since we are asking $$C[0,1]$$ to be sequentially dense in $$(X, \sigma(X,F))$$, even existence of the largest family $$F$$ is not clear, as noted in the comment by @ChristianRemling . So existence of $$F$$ can be considered as a part of the question.

Another update. As pointed out in the comments by @ChristianRemling the term largest is ambiguous. So let me clarify this point. Let $$\mathcal F$$ denote the collection of all families $$F \subset M$$ such that $$C[0,1]$$ is sequentially dense in $$(X, \sigma(X,F))$$. By largest I meant maximal in the partially ordered set $$(\mathcal F, \subset)$$. Now a more precise version of my question has 3 parts:

1. Does there exist a maximal element of $$\mathcal F$$?
2. Does there exist a maximal element of $$\mathcal F$$ which is a superset of the family of all absolutely continuous measures?
3. Can any of these maximal elements (if exists) be described explicitly?
• @ChristianRemling I agree, existence of $F$ is not clear to me at the moment. I thought it would follow from Zorn's lemma, but the point is that the question is about sequential density (added some details to the question). – Skeeve Mar 14 '19 at 17:46
• In fact, it's clear that there is no largest $F$ (in the natural sense that it contains all the others). For example, taking $F$ as any finite collection of Dirac measures works, but we can't take all of them: mathoverflow.net/questions/230028/… – Christian Remling Mar 14 '19 at 21:53
• @ChristianRemling I know that we can't take all Dirac measures. In fact this is exactly the way once proves $F\ne M$ (see also the linked post). But does this really prove non-existence of the largest family $F$? – Skeeve Mar 15 '19 at 7:01
• Say suppose $F\ne M$ is the largest family and we want to prove that it can be extended (to reach a contradiction with the maximality). I know that $C[0,1]$ is sequentially dense in $(X, \sigma(X,F))$ and also in $L^1(X,|\mu|)$, where $\mu \in M \setminus F$, but in general this gives two different approximating sequences. And it is not clear how to conclude from this that $C[0,1]$ is sequentially dense in $(X, \sigma(X, F \cup \{\mu\}))$. – Skeeve Mar 15 '19 at 7:02
• @ChristianRemling I agree that there is no $F_0$ such that $F \subset F_0$ for all $F$, thank you very much for pointing out that we are using different notions of "largest"! I have updated the question to avoid this ambiguity. – Skeeve Mar 15 '19 at 19:06