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There are two questions about measures bothered me a lot.

  1. Given a set X and a countable covering ${U_i}$ of $X$. Suppose that for each i, there is a measure $m_i$ on $U_i$. Is there a very general procedure to define a non-trivial measure $m$ on X by using all of $m_i$´s?

Remark: Each $(U_i,m_i)$ induce a measure $Ind_i$ on $X$ by the obvious way, so by non-trivial, I mean the measures which are not produced from this way. If this covering is finite, then of course we can define a measure via addition of measures. If the intersection $U_{ij} = U_i \cap U_j$ satisfied that $m_k(U_{ij}) = 0$ for $k = i,j$, then we can also use addition of measure. But for infinite covering such that the intersection may have positive measures, I have no ideas.

  1. Given a group $G$ acts on a measure space $(X,m)$ with $m$ is a $G$-invariant non-atomic measure. Is it possible to define a measure on the quotient space $G/X$?

Remark: If $G$ acts on a manifold $X$ equipped with a $G-$invariant measure properly discontinuous, then this can be done.

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  • $\begingroup$ Why not just define the measure of a subset of (I assume you meant, rather than $G/X$) $X/G$ to be the measure of its pullback to $X$? Is the issue that this may not be measureable? $\endgroup$ – LSpice Mar 8 at 18:07
  • $\begingroup$ I want to rule out some weird measures. For example, take X to be real line with the Lebesgue measure, G be the real line regard as translation, then the quotient is a point, if we define the pull-back measure, then I will regard this measure to be weird. $\endgroup$ – BiM Mar 9 at 16:19
  • $\begingroup$ In your example, what would be a "non-weird" measure? We have to assign the point some measure, and $\infty$ seems as well justified to me as any other. $\endgroup$ – LSpice Mar 9 at 18:12
  • $\begingroup$ @LSpice Yes, indeed, In my example, we have to assign some measure to this point, so the infinity is not bad. Sorry about the use of ¨weird¨. But if we compare the original measure with the resulting measure, I have to say that it is indeed a ¨weird¨ measure. So, as in the comment of Nik, I am worried about the case that a nice invariant measure may induce a trivial measure or a non-trivial measure but assign an infinity to some set which we hope to have a positive measure. $\endgroup$ – BiM Mar 9 at 18:57
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The first question seems too broad because there are a lot of things you can do. If they are probability measures you can take a weighted sum such as $\sum \frac{1}{2^n}m_n$. Or you could disjointify by letting $V_n = U_n\setminus(U_1 \cup \cdots \cup U_{n-1})$, restricting each $m_n$ to $V_n$, and summing up. It really depends on what you need this for.

For the second question, a nice way to look at this is by considering the dual action of $G$ on $L^\infty(X)$. If it is finite (or a probability measure), the $G$-invariant measure on $X$ translates to a normal $G$-invariant linear functional on $L^\infty(X)$ (or a normal state), and passing to the quotient corresponds to passing to the subalgebra of $G$-invariant functions in $L^\infty(X)$. This is a von Neumann subalgebra of $L^\infty(X)$, so it is isomorphic to some $L^\infty(Y)$, and you can just restrict the original linear functional on $L^\infty(X)$ to the subalgebra to get a normal linear functional on $L^\infty(Y)$, i.e., a measure on $Y$.

Morally $Y$ is the quotient of $X$ by the $G$-action, but I'm not sure whether you can make this literally true. It seems likely that some kind of measure theoretic pathology would block this from working in general.

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  • $\begingroup$ Really good answers. For the answer of the second question, what do you mean for ¨Morally Y is the quotient¨? when X is a manifold, is Y an object like manifold or orbifold ? And I am still looking for answers for the second question in the case of non-finite measure. $\endgroup$ – BiM Mar 9 at 16:13
  • $\begingroup$ Thank you! "Morally" means "literally true under reasonable extra assumptions". For instance, see Proposition 6.2.4 of my book Mathematical Quantization. Infinite positive measures correspond to weights on von Neumann algebras, and these pass to von Neumann subalgebras just as well, however you might worry about the weight trivializing such that every subset of $Y$ has measure $0$ or $\infty$. $\endgroup$ – Nik Weaver Mar 9 at 17:39

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