Conditional expectation: $E[A \mid B] = B$ and $E[B \mid A] = A$ implies $A = B$

There is no approximation argument (that I know) that allows to deduce the general case from the $L^2$ case. Rather one should go back to the very definition of conditional expectations and note that $A\leqslant B$ almost surely if and only if the events $$ C_x=[A>x\geqslant B] $$ have probability zero for every real number $x$, or for sufficiently many values of $x$. Here is a proof along these lines.

First recall that $\mathrm E(A|B)=B$ means that $\mathrm E(Au(B))=\mathrm E(Bu(B))$ for (at least) every bounded measurable function $u$. Similarly, $\mathrm E(B|A)=A$ means that $\mathrm E(Bv(A))=\mathrm E(Av(A))$ for (at least) every bounded measurable function $v$. In particular, for every real number $x$, $$ \mathrm E(A;B\leqslant x)=\mathrm E(B;B\leqslant x),\qquad \mathrm E(B;A>x)=\mathrm E(A;A>x). $$ Next, introduce the events $$ D_x=[A> x,B> x],\qquad F_x=[A\leqslant x,B\leqslant x]. $$ Then $[B\leqslant x]=C_x\cup F_x$ and $[A>x]=C_x\cup D_x$ and both these unions are disjoint hence $$ \mathrm E(A-B;C_x)+\mathrm E(A-B;F_x)=0=\mathrm E(A-B;C_x)+\mathrm E(A-B;D_x). $$ Summing up these two equalities and using the fact that $\mathrm E(A-B;C_x)\geqslant0$ because $A-B>0$ on $C_x$, one gets $$ \mathrm E(A-B;D_x\cup F_x)\leqslant0. $$ The hypothesis we started from is symmetric with respect to $(A,B)$ hence the conclusion above holds if one replaces $(A,B)$ by $(B,A)$. Then $A-B$ becomes $B-A$ and $D_x$ and $F_x$ do not change. This proves that $\mathrm E(B-A;D_x\cup F_x)\leqslant0$, hence $\mathrm E(A-B;D_x\cup F_x)=0$, which implies $$ \mathrm E(A-B;C_x)=0. $$ But $A>B$ almost surely on $C_x$, hence the random variable $(A-B)\mathbf{1}_{C_x}$ is almost surely nonnegative. This proves that the event $[A>B]\cap C_x=C_x$ has probability zero.

Now, the event $[A>B]$ is the countable union of the events $C_x$ on the rational numbers $x$, hence $[A>B]$ has probability zero. By symmetry, the event $[A<B]$ has probability zero as well, hence $A=B$ almost surely.

This is an interesting question and there are several text books to mention it. The following is copy from A.N. Shiryaev, Problems in Probability, Springer, New York, 2012, p207.