intervals $I_0,I_1,\dots$ with ${\lambda}(I_i)={\sigma}_i$ for a certain signature ${\sigma}$. Further, let us consider a coordinatewise measurable assignment of signatures ${\sigma}^x$ to points $x\in I$ which is constant on the intervals $I_1,I_2,\dots$, and, as before, let $I_0^x,I_1^x,\dots$ be the consecutive subintervals of $I$ with ${\lambda}(I_i^x)={\sigma}_i^x$. Denote by $(X,m)$ the Lebesgue space obtained from the square $(I\times I,{\lambda}\otimes{\lambda})$ by collapsing the sets $\{x\}\times I_i^x,\,i\ge 1,x\in I_0,$ and $I_j\times
I_i^x,\,i,j\ge 1, x\in I_j,$ onto single points. In the same way, let $({\overline}X,{\overline}m)$ be the quotient space of the interval $(I,{\lambda})$ obtained by collapsing the intervals $I_1,I_2,\dots$ onto single points, so that the signature of $({\overline}X,{\overline}m)$ is ${\sigma}$. Then the projection of the square $I\times I$ onto the first coordinate gives rise to a homomorphism from $(X,m)$ to $({\overline}X,{\overline}m)$, and the signatures of the associated conditional measures are precisely ${\sigma}^x$. The claim is that any homomorphism of Lebesgue spaces can be obtained in this way. In particular, if both the quotient measure ${\overline}m$ and all the conditional measures $m_{{\overline}x}$ are purely non-atomic (i.e., have the signature $(1;0,0,\dots)$), then the corresponding quotient map is isomorphic just to the projection of the unit square onto the first coordinate.

Obviously, any homomorphism of Lebesgue spaces gives rise 