Unions and Subsets of Sign-Definite Sets

Sign Conditions Are Conditions on Every Subset

A signed measure can assign positive values to some measurable sets and negative values to others. To call a set positive or negative, it is not enough to inspect the signed measure of the set itself: the sign condition must hold on every measurable part of it. This stronger condition is what makes positive and negative sets useful when constructing larger sets and simplifying their overlaps.

Throughout, let \((X,\mathcal{M},\nu)\) be a measurable space equipped with a signed measure \(\nu\). Here a signed measure is a countably additive set function on \(\mathcal{M}\) with values in \([-\infty,\infty]\), which does not take both \(+\infty\) and \(-\infty\). No finiteness or completeness assumption is imposed. All the set inclusions below are literal set inclusions, and all sign conditions hold on every measurable subset, not merely almost everywhere.

In particular, the test set \(E\) in either definition may be the whole set itself, but the condition is stronger than a condition on \(\nu(P)\) or \(\nu(N)\) alone. A positive set can have positive measure while containing a subset of negative measure only if it fails to be positive; similarly, a negative set cannot contain a subset of positive measure.

Measurable Subsets Inherit the Sign

Proof. Suppose first that \(P\) is positive and \(C\subseteq P\) is measurable. For any measurable \(E\subseteq C\), transitivity of inclusion gives \(E\subseteq P\). The definition of positivity therefore gives \(\nu(E)\ge0\). This holds for every measurable subset \(E\) of \(C\), so \(C\) is positive. If \(N\) is negative and \(C\subseteq N\), then every measurable \(E\subseteq C\) also satisfies \(E\subseteq N\), and hence \(\nu(E)\le0\). Thus \(C\) is negative. \(\square\)

This inheritance rule includes the empty set: its only subset is itself, and \(\nu(\varnothing)=0\) by countable additivity. Thus \(\varnothing\) is both positive and negative. The whole space \(X\), however, is not automatically either; its sign depends on \(\nu\) on all measurable subsets of \(X\).

Countable Unions Preserve a Common Sign

For unions, the challenge is that the sets need not be disjoint. Countable additivity applies directly to disjoint sets, so the useful first step is to replace an overlapping sequence by disjoint pieces. Each piece remains inside one of the original sets, which is precisely where the subset proposition applies.

Proof. We prove the positive assertion first. Since \(\mathcal{M}\) is a sigma-algebra, \(P=\bigcup_{n\ge1}P_n\) is measurable. Define \(D_1=P_1\) and, for \(n\ge2\), \(D_n=P_n\setminus\bigcup_{j=1}^{n-1}P_j\). Each \(D_n\) is measurable, the \(D_n\) are pairwise disjoint, and \(\bigcup_nD_n=P\). Also \(D_n\subseteq P_n\), so the subset proposition implies that every \(D_n\) is positive.

Let \(E\in\mathcal{M}\) be any subset of \(P\). Then the sets \(E\cap D_n\) are measurable and pairwise disjoint, and their union is \(E\). Because \(E\cap D_n\subseteq D_n\) and \(D_n\) is positive, \(\nu(E\cap D_n)\ge0\) for every \(n\). Countable additivity gives \(\nu(E)=\sum_{n\ge1}\nu(E\cap D_n)\). This series consists entirely of nonnegative terms, so its value is nonnegative, including when it is \(+\infty\). Since this holds for every measurable \(E\subseteq P\), \(P\) is positive.

For the negative assertion, set \(D_1=N_1\) and \(D_n=N_n\setminus\bigcup_{j=1}^{n-1}N_j\) for \(n\ge2\). These measurable pieces are disjoint, have union \(N=\bigcup_nN_n\), and are negative by the subset proposition. For every measurable \(E\subseteq N\), each \(E\cap D_n\) has measure at most zero. Countable additivity gives \(\nu(E)=\sum_n\nu(E\cap D_n)\), a sum of nonpositive terms, so \(\nu(E)\le0\), allowing the value \(-\infty\). Thus \(N\) is negative. \(\square\)

The same statement includes finite unions by taking all later sets to be empty, since \(\varnothing\) has both signs. The disjoint pieces also show exactly why overlap causes no difficulty: each point is assigned to its first set in the sequence, and no point is counted twice.

Intersections, Empty Families, and Mixed Signs

There is a useful consequence for intersections, but its measurability and empty-family qualifications matter. The intersection of a nonempty countable family of measurable sets is measurable. If all those sets are positive, their intersection is a measurable subset of any one member, so it is positive by the subset proposition. The same reasoning applies to negative sets. For a nonempty finite family, the result follows in the same way.

By convention, the union of an empty family is \(\varnothing\), and the intersection of an empty family is \(X\). The empty union is both positive and negative. An empty intersection is not automatically sign-definite: it is sign-definite only if \(X\) itself has the relevant sign. This distinction prevents an unjustified use of the subset argument when there is no member of the family to contain the intersection.

Common mistake: adding the values of overlapping sets. In the preceding four-point example, \(\nu(A)=2+1=3\) and \(\nu(B)=1+3=4\). Adding these gives \(7\), but that counts the shared point \(b\), of weight \(1\), twice. The actual union value is \(2+1+3=6\). The correction is to disjointize: \(A\) and \(B\setminus A\) have values \(3\) and \(3\), so countable additivity gives \(3+3=6\). Disjointization is not merely a device for calculating values; it is the step that lets the union proof use countable additivity correctly.

A second caution is that the theorem concerns unions of sets with a common sign. A union of a positive set and a negative set need not be either sign, as the two-point example shows. Nor does the theorem assert closure under arbitrary uncountable unions: an uncountable union of measurable sets need not be measurable, so it may not even be eligible for the definition.

Using the Closure Rules

These properties are useful whenever sign sets overlap or a sign condition must be checked on a larger measurable set. A family of positive sets can be replaced by its union without losing positivity, and the same is true for negative sets. When a calculation requires disjoint sets, the first-occurrence pieces \(D_n\) provide a systematic choice: they are measurable, disjoint, have the same union, and inherit the common sign. Conversely, a proposed sign set can be disproved by finding a single measurable subset with the wrong sign.

In particular, a set's total signed measure by itself never replaces the subset test. The closure results work because every measurable piece has the appropriate sign before countable additivity is applied; the empty-family convention and the disjointization step ensure that this reasoning covers the cases where it is valid.