Background
I am self-studying measure-theoretic probability from the book by Robert Ash.
Definitions
If $X$ is a random variable on $(\Omega, \mathcal{F}, P)$, then the expectation of $X$ is defined by $$ E(X) = \int_\Omega X \; dP, \quad \quad (1)$$ provided the integral exists.
Let $Y$ be an extended random variable on $(\Omega, \mathcal{F}, P)$ and let $E[Y]$ exist. Then for all $\sigma$-fields $\mathcal{G} \subset \mathcal{F}$, there exists an extended random variable $h$ on $(\Omega, \mathcal{G})$ such that: \begin{align*} \int_{\mathcal{G}} Y \, dP = \int_{\mathcal{G}} h \, dP \quad \quad \forall G \in \mathcal{G}. \end{align*} We define the conditional expectation (of $Y$ given the $\sigma$-field $\mathcal{G}$) by $$ E[Y \, | \, \mathcal{G}] := h$$
Question
What justifies the use of the term expectation in the name conditional expectation? After all, the conditional expectation is not defined in terms of an integral, as in (1), but implicitly, as an integrand. Is the answer simply that it behaves like an expectation (because it has properties like monotonicity, linearity, triangle inequality, monotone convergence, dominated convergence, Fatou's Lemma, etc.)?
