Say I have a probability space $(\Omega, \Sigma, \mu)$ and some real-valued random variable $X$ such that the distribution of $X$ is continuous.
Say $\Sigma_{\mathbb{R}}$ is the typical Borel algebra on the reals. Is it always the case that for any non-empty $\sigma \in \Sigma_{\mathbb{R}}$ and for any $b \in \sigma$, we have that:
\begin{align*} \mu(\omega\ |\ X(\omega) \in \sigma) = \\ \int_{a \in \sigma} \mu(\omega\ |\ X(\omega) \in da) = \\ \mu(\omega\ |\ X(\omega) \in \left[\sigma - \{b\}\right]) = \\ \int_{a \in \left[\sigma - \{b\}\right]} \mu(\omega\ |\ X(\omega) \in da) \end{align*}
If this is not the case for all continuous distributions, does this hold if we assume that the distribution of $X$ is Gaussian?
Yes. For $X$ to be continuous means that $P(X=b)=0$ for each $b\in\mathbb R$. The difference between your $P(X\in S)$ and your $P(X\in S\setminus \{ b\})$ is exactly $P(X\in\{b\})=P(X=b)$, which is by assumption zero.