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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.

Write $Y=\mathbb{E}[X\mid \mathcal{G}]$. The conditional expectation is an $L^2$ contraction by the conditional Jensen inequality, so $X,Y$ are square-integrable. The defining integral identity for conditional expectation, first with bounded measurable truncations of $Y$ and then with their $L^2$ limit, givesFor clarity, if $Y^{(K)}=\max (-K,\min (Y,K))$, then $\mathbb{E}[X{Y}^{(K)}]=\mathbb{E}[Y{Y}^{(K)}]$, and the Cauchy-Schwarz inequality permits the limit $K\to \infty$. Expanding the square now proves the equality case for conditional second moments:A nonnegative random variable of zero expected value vanishes almost surely. Therefore $\boxed{X=Y\text{ almost surely}.}$

For the real-valued variables here, take a $\mathcal{G}$-measurable version of $X=\mathbb{E}[Y\mid \mathcal{G}]$. Since the variables are bounded, the conditional expectation defining identity extends to the bounded $\mathcal{G}$-measurable multiplier $X$, givingThe equality case for conditional second moments now givesA nonnegative random variable with zero expectation vanishes with probability one. Thus $X=Y$ as an almost sure equality. The same proof works for square-integrable variables; boundedness is more than is needed.

Write $A=g(X)$ and $B=g(Y)$. The two conditional expectation identities supply a $\sigma (Y)$-measurable version of $B$ and a $\sigma (X)$-measurable version of $A$. Work with those versions, which do not change any almost sure equality. Both lie in L2 space, and the Cauchy-Schwarz inequality makes their product an integrable random variable.

Testing the first conditional expectation identity against $B$, and the second against $A$, givesFor completeness, these tests with unbounded multipliers are valid by first using the bounded truncations $B_m=(-m)\vee (B\wedge m)$ and $A_m=(-m)\vee (A\wedge m)$, then passing to the limit with the Cauchy-Schwarz inequality and convergence in L2. Consequently,A nonnegative random variable with zero expected value vanishes almost surely. Therefore $g(X)=g(Y)$ almost surely. This is mutual conditional expectation rigidity in L2; it also follows from the equality case for conditional second moments.

For a counterexample to equality of the original random variables, let $X$ give probability $1/4$ to each of $-2,-1,1,2$, set $Y=-X$, and use the nonconstant function $g(t)=t^2$. Then $g(X)=g(Y)=X^2$ takes both values $1$ and $4$. It is measurable with respect to both generated sigma-algebras, so both required conditional expectation identities hold. Symmetry gives $\mathbb{E}X=\mathbb{E}Y=0$, while neither random variable can be zero. Thus

The L2-bounded martingale convergence theorem states that a martingale with ${\sup }_n\mathbb{E}|X_n{|}^2<\infty$ converges both almost surely and with convergence in L2 to a square-integrable random variable. Its hypotheses hold because $0<X_n<1$. Thus there is $X_{\infty }\in [0,1]$ almost surely with both kinds of convergence. Alternatively, the martingale convergence theorem gives the almost sure convergence, and the bound $|X_n-X_{\infty }{|}^2\le 1$ gives convergence in L2 by the dominated convergence theorem.^[ref]

The calculation of the conditional expectation of the square is especially informative. The martingale property makes the cross term vanish in the square expansion, soSubtracting this from $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=X_n$ givesTaking expected values and iterating provesBy bounded convergence theorem, $\mathbb{E}[X_{\infty }(1-X_{\infty })]=0$. The random variable inside this expected value is nonnegative, so it is zero almost surely; on $[0,1]$ this forces $X_{\infty }\in \{0,1\}$. Again by bounded convergence theorem, $\mathbb{E}{X}_{\infty }={\lim }_n\mathbb{E}{X}_n=x$. HenceThis is the Bernoulli polarization martingale with update size $1/2$. It also has an exact mean squared error: the terminal representation $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ and the equality case for conditional second moments give