$\sigma$-algebra generated by a weakest topology such that some functions are continuous.

I am wondering about statements like : Let $X$ be a vector space, let $\tau$ be the weakest topology such that some set of functional $F$ are continuous. Let also $\Sigma$ be the Borel $\sigma$-algebra generated by $\tau$. I'm wondering if $\Sigma$ is the weakest $\sigma$-algebra such that all $f\in F$ are measurable.

It is clear that any $f\in F$ is measurable, since any continuous function is measurable. This shows that the weakest $\sigma$-algebra such that $f\in F$ are measurable is contained in $\Sigma$. I'm not sure how to prove the other direction (or if it is true at all).

Not true. Consider the collection of all functions of the form $\chi_{\{x\}}, x \in \mathbb R$. The weak topology generated by this is the discrete topology since each singleton set is open. So $\Sigma$ is the power set of $\mathbb R$. But the smallest $\sigma -$ algebra making each $\chi_{\{x\}}$ measurable is the $\sigma-$ algebra of countable and co-countable sets.

Notation: $\chi_A (y)=1$ if $ y \in A$ and $0$ otherwise.