Computing $E[X\| \mathscr{G}](x)$ from knowledge of which $G$ contain $x$

In Section 34 (page 445) of Billingsley's Probability and Measure (3rd ed.), he says, regarding conditional expectation, that since $E[X \| \mathscr{G}]$ is $\mathscr{G}$-measurable and integrable, $E[X \| \mathscr{G}](x)$ can in principle be calculated just from knowing for each $G \in \mathscr{G}$ whether or not $x \in G$.

Question 1: How can $E[X \| \mathscr{G}](x)$ be calculated just from knowing for each $G \in \mathscr{G}$ whether or not $x \in G$?

My only idea is based on what I'd do in a specific case: Let's suppose $X:\mathbb{R}\to \mathbb{R}$, $\mathscr{G} = \mathscr{B}$ is the Borel $\sigma$-algebra and $\mu$ is Lebesgue measure. If I had access to the knowledge of which $B \in \mathscr{B}$ contain $x$, then in order to compute $E[X \| \mathscr{B}](x)$, I'd search through $\mathscr{B}$ until I found a (non-degenerate) interval $B_{1}$ such that $x \in B_{1}$, at which point I'd compute $E[X|B_{1}] := \frac{1}{\mu(B_{1})} \int_{B_{1}} X \, d\mu$. Then I'd continue searching through $\mathscr{B}$ until I found another interval $B_{2}\subseteq B_{1}$ such that $x \in B_{2}$ and the diameter of $B_2$ is less than half of that of $B_{1}$, at which point I'd compute $E[X|B_{2}]$. Continuing this way, I'd have $B_{1}\supset B_{2} \supset \cdots$ with $\{x\}=\bigcap_{n} B_{n}$ and by the Lebesgue differentiation theorem (LDT),

$$\lim_{n\to \infty} E[X\| B_{n}] = E[X\|\mathscr{B}](x).$$

Question 2: Can this procedure for computing $E[X \| \mathscr{B}]$ be generalized?

Of course, one problem with generalizing is that the LDT doesn't hold in general. It's also not clear how one would choose the $B_{n}$'s.

Update As @CarlMummert pointed out, it doesn't make sense to try to calculate the value of $E[X \| \mathscr{G}](x)$ for any $x$, since conditional expectation is only defined almost everywhere. Thus I am accepting @MichaelGreinecker's answer.

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What is $E[X \; | \; \mathcal{G}]_\omega$? –  Jeff Feb 15 '12 at 14:37
@Jeff It means $E[X\| \mathscr{G}]$ evaluated at the point $\omega$. This is Billingsley's notation. I've changed it to be more standard, though. –  Quinn Culver Feb 15 '12 at 15:30
I don't understand the question. The expectation is only defined up to almost-everywhere equality. If the space is atomless, the value of the conditional expectation at any particular point is meaningless, because the expectation can be changed on any set of measure 0 without affecting anything. Is the question supposed to be about a finite space, or is there some other hypothesis present? –  Carl Mummert Feb 15 '12 at 15:53
@CarlMummert Good point. I'll edit my question. –  Quinn Culver Feb 15 '12 at 20:04
The remark in the Update was already formulated there (a page I thought you knew). –  Did Feb 15 '12 at 20:33
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For question 1: I think one should take the word calculate not too seriously in this context. Billingsley means something much more elementary. In general, explicit calculation with conditional expectations is impossible. No constructive version of the Radon-Nikodym theorem is known.

For notational convenience, denote the mesurable function $E[X \| \mathscr{G}]$ by $f$. For each $r\in\mathbb{R}$, the set $f^{-1}\big(\{r\}\big)$ is measurable. One of these sets contains $x$, so we know the value $f(x)=E[X \| \mathscr{G}](x)$ if we know in which sets in $\mathcal{G}$ the point $x$ lies.

This would not work if $f$ were not measurable. Let $\Omega=\{a,b,c\}$, $\mathcal{G}=\big\{\Omega,\emptyset,\{a,b\},c\big\}$ and $f(a)=f(c)=1$ and $f(b)=0$. Let $x=b$. Then we know that $x$ lies in the measurable sets $\Omega$ and $\{a,b\}$, but both $a$ and $b$ do so. So this is not enough to determine whether $f(x)=0$ or $f(x)=1$.

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Thanks. Notice that, based on a comment of @Jeff, I updated my notation from $E[X\|\mathscr{G}]_{\omega}$ to $E[X\|\mathscr{G}](x)$. It might help future readers if you update the notation in your answer correspondingly. –  Quinn Culver Feb 15 '12 at 15:49
I've updated the notation. –  Michael Greinecker Feb 15 '12 at 15:54
If ${\mathcal G}$ is generated by a finite partition $\{A_1,\ldots,A_n\}$ of the sample space into measurable sets with $P(A_k)>0$ for each $k$, then Billingsley's principle can be taken quite literally, for in this case $E(X\|{\mathcal G})(\omega)=\sum_{k=1}^n E(X|A_k)1_{A_k}(\omega)$---the value of $E(X\|{\mathcal G})(\omega)$ is known once it is known which of the $A_k$s happens to contain $\omega$.
If ${\mathcal G}$ is countably generated, then there is an increasing sequence of $\sigma$-algebras $\{{\mathcal G}_1\subset{\mathcal G}_2\subset\cdots\subset{\mathcal G}$, each generated by a finite partition, with ${\mathcal G}=\sigma\{{\mathcal G}_1,{\mathcal G}_2,\ldots\}$. The martingale convergence theorem then yields $$E(X\|{\mathcal G})(\omega) = \lim_n E(X\| {\mathcal G}_n)(\omega)$$ for almost every sample point $\omega$.