respectively) $$X = [{\mathcal C}\cup ({\mathcal D}\setminus{\mathcal D}_{\mathsf{free}}) ] \sqcup {\mathcal D}_{\mathsf{free}}$$ such that the restriction of the action to $C\cup({\mathcal D}\setminus{\mathcal D}_{\mathsf{free}})$ is conservative and the restriction to ${\mathcal D}_{\mathsf{free}}$ is totally dissipative.

\[cor:DD\] If the action $G{\circlearrowleft}(X,m)$ is free, i.e., ${\mathcal D}={\mathcal D}_{\mathsf{free}}$, then its conservative part coincides with the continual set ${\mathcal C}$, and its dissipative part coincides with the discontinual set ${\mathcal D}$.

\[cor:Poin\] If the measure $m$ is invariant, then ${\mathcal D}_{\mathsf{free}}$ is a null set, and therefore the action is conservative.

The decomposition into the conservative and totally dissipative parts described in the above Theorem is unique. Indeed, let $X=C_1\sqcup
D_1=C_2\sqcup D_2$ be two such decompositions. If they are different, then one of the sets $C_1\cap D_2, C_2\cap D_1$ must be non-empty. Let it be, for instance, $A=C_1\cap D_2$. Then the restriction of the action to $A$ has to be simultaneously conservative (because $A\subset C_1$) and totally dissipative (because $A\subset D_2$), which is impossible.

If the conservativity is replaced with the infinite conservativity, then, generally speaking, the Hopf decomposition as above is not possible. Namely, the action is infinitely conservative on ${\mathcal C}\cup({\mathcal D}\setminus{\mathcal D}_{\mathsf{cofinite}})$ and is completely dissipative on ${\mathcal D}_{\mathsf{free}}$, whereas on 