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Given a measure space $(X,\mathcal{A},\mu)$, what can the range of the measure, $\mu[\mathcal{A}]$, look like? Clearly it can't be an arbitrary subset of $[0,\infty]$ as we know $0\in \mu[\mathcal{A}]$. We also know $\mu[\mathcal{A}]$ has a maximal element ($\mu(X)$).

A bit less trivially, it must also satisfy the following for any $x,y\in \mu[\mathcal{A}]$:

$\exists z\ \{z,x-z,y-z,x+y-z,x+y-2z\}\subseteq\mu[\mathcal{A}]$

($z$ corresponds to the measure of the intersection of the sets $x$ and $y$ correspond to). This for instance tells us $\mu[\mathcal{A}]\ne \{0,1,3\}$.

Additionally, a fact that as far as I can tell is independent of the above comes from measuring the complement of a set: if $x\in\mu[\mathcal{A}]$ then $M-x\in\mu[\mathcal{A}]$, where $M$ is the unique maximal element of $\mu[\mathcal{A}]$ corresponding to $\mu(X)$.

Is any complete characterization known?

edit: for example, at first I thought it might have to be closed, as every natural measure I could think had closed range. But the range of measure on $\mathbb{N}$ generated by $\mu(\{0\})=0.9$, $\mu(\{1\})=0.99$, $\mu(\{2\})=0.999$... has a sequence of elements approaching 1, but does not contain 1 as any singleton set has measure less than 1 and any two-or-more element set has measure greater than 1.

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3 Answers 3

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If it's a finite set, there must be $a_1, \ldots, a_m > 0$ such that $\mu[\mathcal A] = \{ \sum_i a_i x_i : x \in \{0,1\}^m\}$.

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  • $\begingroup$ Thank you, nice to have such a clean characterization in the finite case. I actually wonder if this might somehow follow from my last condition, but that's certainly non-obvious. $\endgroup$ Apr 17, 2020 at 20:00
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It is a well know result by Saks (also generalized by Lyapunov) that if $\mu$ has no atoms, then $\mu$ can take any value in $[0,\mu[X])$. If $\mu$ has atoms, the range of $\mu$ can get a little weird.

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  • $\begingroup$ According to Wikipedia, that theorem is due to Sierpinski. $\endgroup$ Oct 15, 2020 at 13:20
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The range of a measure is either $[0,\infty]$, or there exists $a\in[0,\infty)$ and $S\subset[0,\infty]$ such that $$ \mu[\mathcal{A}] = [0,a] + \left\{\sum_{x\in X} x \mid X\subset S\right\} $$ where we interpret the sum of two sets $A+B = \{a+b \mid a\in A, b\in B\}$.

Non-atomic case: As Oliver Diaz has pointed out, it is a well-known result that if $\mu$ has no atoms, then $\mu$ can take any value in $[0,\mu(X)]$. See https://en.wikipedia.org/wiki/Atom_%28measure_theory%29.

Purely atomic case: Suppose $\mu$ is purely atomic. Let $\mathcal{B}\subset\mathcal{A}$ be the set of atomic sets, i.e. for all $B\in\mathcal{B}$, $\mu(B)\ne 0$ but any measurable subset of $B$ has measure $0$ or has the same measure as $B$, and let $\mu(\mathcal{B})$ be the set of measures of atomic sets. For any $B_1,B_2\in\mathcal{B}$, it's straightforward to see that either their intersection has measure $0$ or their symmetric difference has measure $0$. Furthermore, we can take countable unions of them and still get a measurable set, and one can see that any measurable set with finite measure must be such a countable union plus a null set. Hence we have that if $\mu$ is purely atomic and $\mathcal{B}$ is the set of atoms, then for any measurable $A$ either $\mu(A) = \infty$ or $$ \mu(A) = \sum_{b\in\mu(\mathcal{B})}^\infty b x_b $$ where $x_b:\mu(\mathcal{B})\to\{0,1\}$ has countable support.

General $\sigma$-finite case: Any $\sigma$-finite measure $\mu$ can be decomposed uniquely as $\mu = \nu+\alpha$ where $\nu$ is non-atomic and $\alpha$ is purely atomic. Letting $C$ be the set of measures of atoms of $\alpha$, we have that for any $A\in\mathcal{A}$, either $\mu(A)=\infty$ or $$ \mu(A) = t + \sum_{c\in C} c x_c $$ where $x_c : C\rightarrow \{0,1\}$ has countable support and $t\in[0,\nu(X)]$. Given any such $x$ and $t$, one can construct a set with the corresponding measure, so this is the full range of $\mu$.

Thus, given any $a\in[0,\infty]$ and set $S\subset [0,\infty]$, there exists a measure space $(X,\mu,\mathcal{A})$ such that $$ \mu[A] = [0,a] + \left\{\sum_{x\in X} x \mid X\subset S\right\} $$ and all $\sigma$-finite measures have ranges of this form.

General case: Non-$\sigma$-finite measures are often weird and messy, but one can show that their ranges are the same shape of sets. Suppose $(X,\mathcal{A},\mu)$ is not $\sigma$-finite and the range of $\mu$ is not $[0,\infty]$. Then we can let $a = \sup\{x\in\mathbb{R}\mid \exists A\in\mathcal{A}\text{ non-atomic with }\mu(A)=x\}$. Let $X_a\in\mathcal{A}$ be a set with measure $a$. Then any finite-measure non-null set that is disjoint from $X_a$ must contain an atom, which we can show as follows: Let $Y\in\mathcal{A}$ have finite measure and $Y\cap X_a = \emptyset$. If $Y$ contains no atoms, then $(Y,\mathcal{A}\cap P(Y), \mu)$ is a non-atomic measure space, hence $[0,\mu(Y)]\subset\mu[A]$. But, since $Y$ is disjoint from $X_a$, for any measurable subset $Z\subset Y$, $\mu(X_a\cup Z) = a+\mu(Z) > a$, which cannot happen by the definition of $a$, as it would imply $[0,\mu(Y)+a]\subset\mu[\mathcal{A}]$. This furthermore implies that any set with finite non-zero measure disjoint from $X_a$ can be written as a countable union of atoms. Hence, if $C$ is the set of all possible measures of atoms disjoint from $X_a$, the measure of any finite-measure set can be written as $$ t+\sum_{c\in C}c x_c $$ where $t\in[0,a]$ and $x_c:C\to\{0,1\}$ has countable support. Hence any measure's range has the form $$ [0,a] + \left\{\sum_{x\in X} x \mid X\subset S\right\} $$ for some $a\in[0,\infty]$ and $S\subset [0,\infty]$.

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    $\begingroup$ Thank you so much! This has been bothering me a while, this is perfect! $\endgroup$ Oct 16, 2020 at 2:34

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