If $(X, \Sigma)$ is a measurable space and $Y$ a topological space, and we assume that $\Sigma$ includes the Borel algebra, then $f : X \to Y$ continuous implies $f : X \to Y$ measurable. I believe that there should be a converse under certain conditions on $X$: that if all continuous functions from $X$ to $Y$ are measurable, then $\Sigma$ must contain the Borel algebra. Perhaps $X$ must be, e.g., Hausdorff and locally compact.
Is there any result of this kind? Any suggestions for resources that discuss these kind of issues would be helpful.
My motivation for asking this question is a throwaway line in Douglas' book Banach algebra techniques in operator theory, on page 93, on which he says that a measure $\mu$ on a compact set $\Lambda \subseteq \mathbb{C}$ is a finite, positive, regular Borel measure iff the inclusion $C(\Lambda) \to L^{\infty}(\Lambda, \mu)$ is an isometry. I am trying to prove this; if anyone has sources that prove this statement directly, those would also be appreciated.
You definitely need more assumptions. I can give a counter example: Let us regard $X$ with a sigma-algebra $\Sigma_{X}$ and a topology $\mathcal{T}_{X}$. Then we choose $Y = \{0\}$ with the sigma-algebra $\Sigma_Y = \{\emptyset,\{0\}\}$ and the topology $\mathcal{T}_Y = \{\emptyset,\{0\}\}$. Then there exists only one function from $X$ to $Y$ and this function is continuous and measurable, regardless of the topology and the sigma-algebra of $X$.