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If $X$ is an integrable random variable and $Z$ is a finite, strictly positive $\mathcal{G}$-measurable random variable, thenIndeed, $0\le Z{1}_{\{|X|\ge Z\}}\le |X|$. Monotonicity of conditional expectation and the measurable-factor rule for conditional expectation giveDivision by the strictly positive $Z$ proves the claim. Only the displayed product needs to be an integrable random variable; neither $Z$ nor $|X|/Z$ needs to have finite expected value. The measurable threshold represents information already available before asking how likely the remaining exceedance is.

Let $A,B$ belong to L2 space, and suppose $B=\mathbb{E}[A\mid \mathcal{H}]$ and $A=\mathbb{E}[B\mid \mathcal{G}]$ almost surely, for two possibly nonnested sigma-algebras. Then $A=B$ almost surely. No independence or inclusion between the sigma-algebras is required.

The identities supply $\mathcal{H}$-measurable and $\mathcal{G}$-measurable versions of $B$ and $A$. The measurable-factor rule for conditional expectation and Cauchy-Schwarz inequality giveExpanding the square shows $\mathbb{E}[(A-B{)}^2]=0$, and hence the required almost sure equality. Equivalently, each conditional expectation is an orthogonal projection in L2 space, and the two projection identities force equality in the equality case for conditional second moments.

Equality after applying a function need not identify its original input. For instance, on four equally likely outcomes let $X$ take $-2,-1,1,2$ and set $Y=-X$. Their squares are the same nonconstant random variable, even though $X$ and $Y$ differ everywhere. Injectivity of the function would rule out this loss of information.