Basic question on (regular) conditional probablities.

Question

Let $(\Omega,\mathcal{A},P)$ be a measure space and $P(A|\mathcal{C})$ be the conditional probability of $A$ under $\mathcal{C}$. A regular conditional probablity is a conditional probablity that is also a measure for every fixed $\omega\in\Omega$. Let $(A_{n})_{n\in\mathbb{N}}\in\mathcal{A}$ with $A_{i}\cap A_{j}=\emptyset\ \forall i\neq j $. We then know that: \begin{equation*} P(\bigcup \limits_{n\in\mathbb{N}}A_{n}) = \sum_{n\in\mathbb{N}} P(A|\mathcal{C}) \quad P\text{-a.s.} \end{equation*} But somehow this is not a measure for fixed $\omega\in\Omega$? At least that's what I've read in books but the explanations were very short unfortunately. Let's say $(A_{n})_{n\in\mathbb{N}}\subseteq\mathcal{A}$. Then: \begin{equation*} P(\bigcup \limits_{n\in\mathbb{N}}A_{n}) = \sum_{n\in\mathbb{N}} P(A|\mathcal{C}) \quad \text{ for all } \omega\in N^{c}, \text{ for some } N\in\mathcal{C} \text{ with } P(N)=0. \end{equation*} Then for all $\omega\in\Omega$ it holds: \begin{equation*} 1_{N^{c}}P(\bigcup \limits_{n\in\mathbb{N}}A_{n}^{i}) = 1_{N_{i}^{c}}\sum_{n\in\mathbb{N}} P(A|\mathcal{C}), \end{equation*} ($1_{N}$ is an indicator function). We could do this even for uncountable many sequences. Then wouldn't this be a measure for fixed $\omega\in\Omega$ if we let $A$ run? Thanks for your help, I'v been thinking about this for days but can't wrap my head around it.