Joint-measure of POVM's I feel disturbed by this question: Suppose $A$ and $B$ are POVM's with respective $\sigma$-algebras $\mathcal{F}_A$ and $\mathcal{F}_B$ and outcome spaces $\Omega_A$ and $\Omega_B$. Then why can't I take the following POVM $Z = A\cdot B$ defined on $\mathcal{F}_A \otimes \mathcal{F}_B$ as the joint measure? I mean it fullfills that 
$$
Z(U, \Omega_B) = A(U) \cdot B(\Omega_B) = A(U), \quad \forall U \in \mathcal{F}_A
$$
and the same for $B$. And I guess you can define it to be $\sigma$-additive, at least in finite dimension of the outcome spaces? 
 A: What about positivity? The product of bounded positive operators is positive if they commute (see proof below), otherwise there is no guarantee. If your initial POVMs are not compatible, in general, the operators of the final candidate POVM is not made of positive operators and thus they do not define a POVM. 
Proposition. If  $A,B \geq 0$ where $A,B :\cal H \to \cal H$ are bounded with $\cal H$ Hilbert space and $AB=BA$ then $AB \geq 0$.
PROOF.
It is known that if $A\geq 0$ is bounded, then there is a unique positive bounded operator, $\sqrt{A}$, such that $\sqrt{A}^2 =A$. Moreover that operator commutes with all bounded operators commuting with $A$. In the present case $AB= \sqrt{A}\sqrt{A}B$ and, since  $A$ and $B$ commute, $AB= \sqrt{A}\sqrt{A}B= \sqrt{A}B \sqrt{A}$. Finally, using the fact that a bounded positive opertor is self-adjoint,
$$\langle x, AB  x \rangle= \langle x \sqrt{A}B \sqrt{A} x \rangle = 
\langle \sqrt{A} x , B \sqrt{A} x \rangle = \langle y , B y \rangle \geq 0$$
because $B\geq 0$. Since $x\in \cal H$ is arbitrary, it implies that $AB \geq 0$. QED
