I was looking on some old notes on probability theory and I found the following
"Let $\left(\Omega, \Sigma, P\right)$ be some probability space, $\Sigma_1, \Sigma_2 \subseteq \Sigma$ be two sigma algebras and $X$ be some random variable. Prove or disprove the following: $ \textrm E\left[ \textrm E\left[ X | \Sigma_1\right] | \Sigma_2 \right] = \textrm E\left[ \textrm E\left[ X | \Sigma_2\right] | \Sigma_1 \right] = \textrm E\left[ X | \Sigma_1 \cap \Sigma_2 \right] $"
I think this is not in general true, unless if we add the $\Sigma_1 \cap \Sigma_2$-measurability for both $\textrm E\left[ \textrm E\left[ X | \Sigma_1\right] | \Sigma_2 \right] $ and $ \textrm E\left[ \textrm E\left[ X | \Sigma_2\right] | \Sigma_1 \right] $ (which tranfsorms the question to a standard property of the conditional expectation). In the case $\Sigma_1 \cap \Sigma_2 = \left\{ \emptyset, \Omega \right\}$ then $ \textrm E\left[ X | \Sigma_1 \cap \Sigma_2 \right] $ should be constant, which is not necessarily the case with the other two conditional expectations.
Counterexample: When $\left(\Omega= [0,1) , \Sigma = \mathcal M_\lambda, P = \lambda\right)$ ($\lambda$ being the Lebesgue measure) define $ X : \Omega = [0,1) \longmapsto \mathbb R : t \longmapsto X(t) = \pi \sin(3\pi t)$, and $\Sigma_1 = \left\{ \emptyset, \Omega, \left[0, 1/3\right), \left[1/3, 1\right) \right\} $, $\Sigma_2 = \left\{ \emptyset, \Omega, \left[0, 2/3\right), \left[2/3, 1\right) \right\} $
Is this correct?
Yes, you are correct. The quoted statement is false, and your counterexample is valid.