Let $A$ and $B$ be two unbounded self-adjoint (+ positive if needed) operators with domains $D(A)$ and $D(B)$ respectively. By the spectral theorem, we know that $A=\int_{\mathbb{R}}\lambda dE$ and

$B=\int_{\mathbb{R}}\mu dF$ for some unique spectral measures $E$ and $F$. The commonly known definition of strong commutativity of $A$ and $B$ states that this is the case if $E$ commutes with $F$.

Now, if $A$ commutes strongly with $B$, then a new two-parameter measure, noted $G$, may be defined as the product of $E$ and $F$ so that

$$\overline{BA}=\overline{AB}=\int_{\mathbb{R}^2}\lambda\mu dG.$$

My question is: If $A$ and $B$ are two unbounded self-adjoint (also positive if needed) such that $AB$ is self-adjoint, then does it follow that $A$ commutes strongly with $B$?

Recall that if $B\in B(H)$, then the above question is true.