# What can the range of a measure be?

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.

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

• 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. Apr 17, 2020 at 20:00

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.

• According to Wikipedia, that theorem is due to Sierpinski. Oct 15, 2020 at 13:20

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 $$S\subset [0,\infty]$$.

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