Why the Sign of the Whole Set Is Not Enough
A signed measure can assign positive measure to a set that contains a measurable subset of negative measure. For that reason, the phrase “positive set” does not mean merely a set whose total signed measure is nonnegative. It describes a stronger property: every measurable part of the set must have nonnegative signed measure. The corresponding definition for a negative set requires every measurable part to have nonpositive signed measure.
Throughout this tutorial, \((X,\mathcal{M})\) is a measurable space and \(\nu\) is a signed measure on it, with the usual convention that \(\nu\) is countably additive and does not take both \(+\infty\) and \(-\infty\) as values. No finiteness assumption is needed for the definitions below. In the density examples, \((X,\mathcal{M},\mu)\) is a measure space, \(h:X\to\mathbb{R}\) is measurable and integrable, and \(\nu(A)=\int_A h\,d\mu\). In those examples \(\nu\) is finite. All subset conditions refer to measurable subsets, and inequalities for \(\nu\) are inequalities of extended real numbers.
Taking \(A=P\) or \(A=N\) shows that a positive set has nonnegative measure and a negative set has nonpositive measure. Those conclusions are necessary, but they are not sufficient: the definition tests every measurable subset as well as the set itself. The requirement also allows zero-measure pieces and subsets of measure zero; “positive” and “negative” refer to the direction of the inequalities, not to strict inequalities.
Finite Atomic Measures: Check Each Atom
On a finite space with every subset measurable, every subset can be assembled from singletons. This makes it possible to reduce the definition to a finite list of weights.
Proof. Suppose first that \(P\) is positive. For each \(x\in P\), the singleton \(\{x\}\) is a measurable subset of \(P\), so \(w_x=\nu(\{x\})\ge0\). Conversely, suppose every \(w_x\ge0\) for \(x\in P\). Any measurable subset \(A\subseteq P\) is finite, and finite additivity gives \(\nu(A)=\sum_{x\in A}w_x\ge0\). Thus \(P\) is positive. For the negative assertion, if \(N\) is negative, applying its definition to each singleton gives \(w_x\le0\) for \(x\in N\). If all those weights are nonpositive, then every subset \(A\subseteq N\) has \(\nu(A)=\sum_{x\in A}w_x\le0\), so \(N\) is negative. The empty set satisfies both directions, with an empty sum equal to zero. \(\square\)
Worked Example: Testing Sets in a Four-Point Space
Let \(X=\{a,b,c,d\}\), let every subset be measurable, and assign weights \(w_a=3\), \(w_b=-2\), \(w_c=0\), and \(w_d=1\). Thus, for every \(A\subseteq X\), \(\nu(A)=\sum_{x\in A}w_x\). Test \(P=\{a,c,d\}\), \(N=\{b,c\}\), and \(X\).
The weights in \(P\) are \(3,0,1\), all nonnegative, so the proposition says \(P\) is positive. To check the definition directly, its eight subsets have measures \(0\) for \(\varnothing\), \(3\) for \(\{a\}\), \(0\) for \(\{c\}\), \(1\) for \(\{d\}\), \(3\) for \(\{a,c\}\), \(4\) for \(\{a,d\}\), \(1\) for \(\{c,d\}\), and \(4\) for \(\{a,c,d\}\). Each is nonnegative.
The weights in \(N\) are \(-2,0\), both nonpositive, so \(N\) is negative. Its four subsets have measures \(0\) for \(\varnothing\), \(-2\) for \(\{b\}\), \(0\) for \(\{c\}\), and \(-2\) for \(\{b,c\}\), each nonpositive. For the whole space, \(\nu(X)=3-2+0+1=2\), but \(X\) is not positive because the subset \(\{b\}\) has measure \(-2\); it is not negative because \(\{a\}\) has measure \(3\). Check: the sum of the four singleton weights is \(3+(-2)+0+1=2\), matching \(\nu(X)\).
The Exact Test for an Integrable Density
For a signed measure given by an integrable density, the subset condition has a direct pointwise interpretation. Here \((X,\mathcal{M},\mu)\) is any measure space, \(h:X\to\mathbb{R}\) is measurable with \(\int_X|h|\,d\mu<\infty\), and \(\nu(A)=\int_A h\,d\mu\) for \(A\in\mathcal{M}\). The phrase \(h\ge0\) \(\mu\)-a.e. on \(P\) means that \(\{x\in P:h(x)<0\}\) has \(\mu\)-measure zero; the analogous meaning applies to \(h\le0\) \(\mu\)-a.e.
Proof. We prove both directions for positive sets. If \(h\ge0\) \(\mu\)-a.e. on \(P\), then for every measurable \(A\subseteq P\), \(h\ge0\) \(\mu\)-a.e. on \(A\). Therefore \(\nu(A)=\int_A h\,d\mu\ge0\), by the sign property of the Lebesgue integral. This holds for every such \(A\), so \(P\) is positive.
Conversely, suppose \(P\) is positive. If \(h\ge0\) \(\mu\)-a.e. on \(P\) were false, then the measurable set \(B=P\cap\{h<0\}\) would have positive \(\mu\)-measure. Since \[ B=\bigcup_{n=1}^{\infty}\bigl(P\cap\{h\le-1/n\}\bigr), \] at least one set \(C=P\cap\{h\le-1/n\}\) would have positive measure: if each had measure zero, countable subadditivity would give \(\mu(B)=0\). Since \(|h|\) is integrable and \(|h|\ge1/n\) on \(C\), we have \(\mu(C)<\infty\); otherwise \(\int_C|h|\,d\mu\ge\mu(C)/n=\infty\), contradicting integrability. On \(C\), \(h\le-1/n\), and hence \[ \nu(C)=\int_C h\,d\mu\le-\frac{1}{n}\mu(C)<0. \] This contradicts positivity of \(P\), because \(C\subseteq P\). Thus \(h\ge0\) \(\mu\)-a.e. on \(P\).
For negative sets, if \(h\le0\) \(\mu\)-a.e. on \(N\), then every measurable \(A\subseteq N\) has \(\nu(A)=\int_Ah\,d\mu\le0\). Conversely, if \(N\) is negative but \(h\le0\) fails on a set of positive measure in \(N\), then \(N\cap\{h>0\}\) has positive measure. It is the union over \(n\ge1\) of \(C_n=N\cap\{h\ge1/n\}\), so some \(C_n\) has positive measure. Integrability of \(|h|\) implies \(\mu(C_n)<\infty\), and \(\nu(C_n)\ge\mu(C_n)/n>0\), contradicting that \(N\) is negative. Both equivalences follow. \(\square\)
Worked Example: A Density That Changes Sign
On \(([-1,1],\mathcal{L},m)\), define \(\nu(A)=\int_A x\,dm\), where \(m\) is Lebesgue measure. Test \(P=[0,1]\), \(N=[-1,0]\), and \(E=[-1,1]\).
The density \(h(x)=x\) is nonnegative on \(P\), so the theorem shows that \(P\) is positive. More explicitly, every measurable \(A\subseteq P\) satisfies \(\nu(A)=\int_Ax\,dm\ge0\). Likewise, \(x\le0\) on \(N\), so \(N\) is negative and every measurable \(A\subseteq N\) satisfies \(\nu(A)\le0\). The shared point \(0\) contributes zero and causes no difficulty.
The total measure of \(E\) is \[ \nu(E)=\int_{-1}^{1}x\,dx=\left[\frac{x^2}{2}\right]_{-1}^{1}=\frac12-\frac12=0. \] But \(E\) is neither positive nor negative: its measurable subset \([-1,0]\) has measure \(\int_{-1}^{0}x\,dx=-1/2\), while its subset \([0,1]\) has measure \(\int_0^1x\,dx=1/2\). Check: adding those subset measures gives \(-1/2+1/2=0=\nu(E)\), while their opposite signs rule out either sign condition on \(E\).
Worked Example: Testing Every Subset Through Its Density
On \((\mathbb{R},\mathcal{L},m)\), let \[ h=2\chi_{(0,1)}-\chi_{(1,2)},\qquad \nu(A)=\int_A h\,dm. \] Determine whether \(P=(0,1)\) is positive, whether \(N=(1,2)\) is negative, and whether \(E=(0,2)\) is positive or negative.
The density is \(2\) on \(P\), so for every measurable \(A\subseteq P\), \(\nu(A)=2m(A)\ge0\); hence \(P\) is positive. In particular, \(\nu(P)=2m((0,1))=2\cdot1=2\). On \(N\), the density is \(-1\), so for every measurable \(A\subseteq N\), \(\nu(A)=-m(A)\le0\); hence \(N\) is negative, and \(\nu(N)=-1\cdot1=-1\). The endpoints \(0,1,2\) are not in these open intervals, and changing values at endpoints would not affect these Lebesgue integrals.
The total measure of \(E\) is \(\nu(E)=2m((0,1))-m((1,2))=2\cdot1-1\cdot1=1\). Nevertheless, \(E\) is not positive, since its measurable subset \(N\) has measure \(-1\). It is not negative, since its subset \(P\) has measure \(2\). Check: the two pieces give \(\nu(E)=\nu(P)+\nu(N)=2+(-1)=1\), consistent with the integral over \(E\).
A Common Mistake and a Practical Test
A common mistake is to test only the whole set. For instance, in the preceding example one might calculate \(\nu((0,2))=1\) and conclude that \((0,2)\) is positive. That calculation establishes only that the whole set has positive measure; the subset \((1,2)\) has measure \(-1\), so the required condition fails. The correction is to verify the sign on every measurable subset, or, when a density is available, to use the theorem and check the density's sign almost everywhere on the set.
The distinction between pointwise sign and almost-everywhere sign matters. For density measures, a density may have either sign on a \(\mu\)-null subset without changing any of the integrals that define \(\nu\). Thus the exact criterion is \(h\ge0\) \(\mu\)-a.e. on a positive set, not necessarily \(h(x)\ge0\) at every point. The theorem does not assert that \(\nu(P)>0\): a positive set may have measure zero, for example if its density vanishes almost everywhere there.
For a finite atomic measure, the practical test is to inspect each singleton weight in the candidate set. For an integrable density, the practical test is to inspect the sign of that density almost everywhere on the candidate set. In both cases the test works because it controls the measure of every measurable subset, which is the defining requirement—not merely the value of the measure on the candidate set itself.
Check Your Understanding
For each question, distinguish the sign of a set's total measure from the sign condition on all its measurable subsets.
- On a finite measurable space with singleton weights \(4,-1,0\), which subsets are positive according to the atomic criterion?
- If \(\nu(A)=\int_A h\,d\mu\) with \(h\) integrable, state the necessary and sufficient condition for \(P\) to be positive.
- For \(h(x)=x-2\) on \([0,3]\) with Lebesgue measure, is \([0,3]\) positive, negative, or neither? Give a measurable subset that verifies your answer.
- Why does \(\nu(P)\ge0\) alone not prove that \(P\) is positive?
- Can a set be both positive and negative? Give an example or justify why not.