Condition implying that $E(X|\mathcal{C}_1)=E(X|\mathcal{C}_2)$ when $\mathcal{C}_1\subseteq\mathcal{C}_2$ I have the following corollary in my notes but I don't see how it follows from law of total probability:
Corollary. Assume that $\mathcal{C}_1\subseteq\mathcal{C}_2\subseteq\mathcal{F}$ are sub $\sigma$-fields and $E(X)$ exists. If a version of $E(X|\mathcal{C}_2)$ is $\mathcal{C}_1/\mathcal{B}^1$-measurable then $E(X|\mathcal{C}_2)$ is also a version of $E(X|\mathcal{C}_1)$.
In my notes the law of total probability stands as follows:
Theorem. If $\mathcal{C}_1\subseteq\mathcal{C}_2\subseteq\mathcal{F}$ are sub $\sigma$-fields and $E(X)$ exists, then $E(X|\mathcal{C}_1)$ is version of $E(E(X|\mathcal{C}_2)|\mathcal{C}_1)$.
Any hint is appreciated.
 A: What is $E[X|\mathcal C_1]$?
It is any random variable $Z$ that satisfies


*

*$Z$ is $\mathcal C_1$-measurable.

*$$\int_{C_1} E[X|\mathcal C_1] d\mathbb P = \int_{C_1} Z d\mathbb P$$
$$\forall C_1 \in \mathcal C_1$$
Is $E[X|\mathcal C_2]$ one of those $Z$'s?


*

*$E[X|\mathcal C_2]$ is $\mathcal C_1$-measurable by assumption. (*)


2.
$$LHS = \int_{C_1} E[X|\mathcal C_1] d\mathbb P$$
$$ = \int_{\Omega} 1_{C_1} E[X|\mathcal C_1] d\mathbb P$$
$$ = \int_{\Omega} E[X1_{C_1}|\mathcal C_1] d\mathbb P$$
$$ = E[E[X1_{C_1}|\mathcal C_1]]$$
$$ = E[X1_{C_1}]$$
$$RHS = \int_{C_1} Z d\mathbb P$$
$$ = \int_{C_1} E[X|\mathcal C_2] d\mathbb P$$
$$ = \int_{\Omega} 1_{C_1} E[X|\mathcal C_2] d\mathbb P$$
$$ = E[1_{C_1}E[X|\mathcal C_2]]$$
$$ = E[E[X1_{C_1}|\mathcal C_2]]$$
$$ = E[X1_{C_1}]$$
I believe #2 does not make use of #1 unlike below

This is the formal way to go about this. I believe there's an intuitive way of thinking about it.

And of course there's the shortcut
$$E[X|\mathcal C_1] = E[E[X|\mathcal C_2]|\mathcal C_1]$$
$\because E[X|\mathcal C_2]$ is given to be $\mathcal C_1$-measurable $(*)$, we have
$$RHS = E[E[X|\mathcal C_2]|\mathcal C_1] \stackrel{(*)}{=} E[X|\mathcal C_2] E[1|\mathcal C_1] = E[X|\mathcal C_2] (1)$$

$(*)$ This is critical. There's no reason for $E[X|\mathcal C_2]$ to surely be $\mathcal C_1$-measurable otherwise because $\mathcal C_2$ is the bigger $\sigma-$field.
