Given $\Omega = \{ a,b,c\}$, I want to construct an example where $E(E(X \mid \mathcal{F}_1) \mid \mathcal{F}_2) \neq E(E(X \mid \mathcal{F}_2) \mid \mathcal{F}_1)$. This is from Durrett's probability theory and examples.
Let $\mathcal{F}_1 = \sigma(\{a\}) = \{ \{a\}, \{b,c\}, \{a,b,c\}, \emptyset\}$ and let $\mathcal{F}_2 = \sigma(\{ c\}) = \{\{c\}, \{a,b\}, \{a,b,c\}, \emptyset \}$. Let $X(a) = X(c) = 0$ and $X(b) = 1$, and $P(a) = P(b)= P(c) = 1/3$. Then in the text, we apparently get
$$E(X\mid \mathcal{F}_1)(a) = 0,\quad E(X\mid \mathcal{F}_1)(b) = 1/2, \quad E(X\mid \mathcal{F}_1)(c) = 1/2 $$
$$E(E(X\mid \mathcal{F}_1)\mid \mathcal{F}_2)(a) = 1/4, \quad E(E(X\mid \mathcal{F}_1)\mid \mathcal{F}_2)(b) = 1/4, \quad E(E(X\mid \mathcal{F}_1)\mid \mathcal{F}_2)(c) = 1/2$$
I think what the book did here is use $E(X \mid \mathcal{F}) = {E(X ; \Omega_i) \over P(\Omega_i)}$ on $\Omega_i$, so for $\{b,c\}$,
$$E(X\mid \mathcal{F}_1) (\{b,c\}) = {E(X ; \{b,c\}) \over P(\{b,c\})} = {1/3 \over 2/3} = {1 \over 2}$$
But how are $E(X\mid \mathcal{F}_1)(b) = 1/2$, $E(X\mid \mathcal{F}_1)(c) = 1/2$ when $\{b\}$ and $\{c\}$ are not even in $\mathcal{F}_1$. Even if we assume $E(X\mid \mathcal{F}_1)(c)$ can be computed, shouldn't $E(X\mid \mathcal{F}_1)(c) = 0$ since $E(X; \{c\}) = 0$?
$E(X|\mathcal{F}_1)$ is by definition $\mathcal{F}_1$- measurable, so we have
$E(X|\mathcal{F}_1)(\omega)=\alpha {1}_{\{a\}}(\omega) + \beta 1_{\{b,c\}}(\omega)$ for all $\omega \in \Omega$ and some $\alpha, \beta\in \mathbb{R}$. Moreover, it easily follows from the definition of the conditional expectation that
$\beta=\frac{E(X \cdot 1_{\{b,c\}} )}{P(\{b,c\})}=\frac{1\cdot \frac{1}{3}+0\cdot \frac{1}{3}}{2/3}=\frac{1}{2}$.
Therefore, $E(X|\mathcal{F}_1)(b)=E(X|\mathcal{F}_1)(c)=\beta=\frac{1}{2}$.