I am dipping my toes into measure-theoretic probability theory for the first time, and I've come across a very basic question.
Suppose I have a $\sigma$-algebra defined by
$$\mathcal{F} = \{\varnothing, \{a\}, \{b, c\}, \Omega \}$$
and a probability measure $\mathbb{P}(\{a\}) = \mathbb{P}(\{b\}) = \mathbb{P}(\{c\}) = \frac{1}{3}$. Furthermore, define a random variable
$$X = \begin{cases}a \to 1 \\ b \to 2 \\ c \to 3 \end{cases}$$
Then by the definition of the conditional expectation, we should have
$$\int_{\{a\}} \mathbb{E}\left[ X \mid \mathcal{F} \right] \text{d}\mathbb{P} = \int_{\{a\}} X \text{d}\mathbb{P}$$
How do we take the integral on the RHS? Is it correct to say
$$\int_{\{a\}} X \text{d}\mathbb{P} = X(a) \mathbb{P}(\{a\}) = \frac{1}{3}$$
$X$ is a simple function, therefore the expression $\int_{A}X\mathrm{d}P$ is for any measurable set $A$ defined via: $$\sum x_i P(X^{-1}(x_i) \cap A),$$ wherein the $x_i$ are the distinct values the r.v. $X$ takes. In essence this leads to the same expression as you wrote.