Finite Unions of Nearly Minimizing Sets

Combining Sets Near the Infimum

For a positive measure, enlarging a set cannot decrease its measure. A signed measure has no such monotonicity: adding a region with negative measure can lower the value, while adding a region with positive measure can raise it. The useful replacement is an exact identity involving both the union and the intersection. It lets us control the value of a finite union of sets whose signed measures are close to the infimum.

Throughout, \((X,\mathcal{M})\) is a measurable space and \(\nu\) is a finite, real-valued signed measure on it. Thus \(\nu\) is countably additive on \(\mathcal{M}\). We use the standard lower-bound property for a finite signed measure: there is a real number \(L\) such that \(\nu(E)\ge L\) for every \(E\in\mathcal{M}\). Define

Because \(\nu(\varnothing)=0\), we have \(\alpha\le 0\); the lower bound gives \(\alpha\ge L\). Hence \(\alpha\) is a finite real number. By the definition of infimum, for every \(\varepsilon>0\) there is a measurable set \(E\) with \(\nu(E)<\alpha+\varepsilon\). Such a set will be called nearly minimizing at tolerance \(\varepsilon\).

The Union–Intersection Identity

Proof. The sets \(A\setminus B\), \(B\setminus A\), and \(A\cap B\) are measurable and pairwise disjoint. Their union is \(A\cup B\). Finite additivity, which follows from countable additivity by adjoining empty sets, gives \(\nu(A\cup B)=\nu(A\setminus B)+\nu(B\setminus A)+\nu(A\cap B)\). Also, \(A\) is the disjoint union of \(A\setminus B\) and \(A\cap B\), while \(B\) is the disjoint union of \(B\setminus A\) and \(A\cap B\). Therefore \(\nu(A)+\nu(B)=\nu(A\setminus B)+\nu(B\setminus A)+2\nu(A\cap B)\). Comparing these equalities proves the identity. The argument also applies if either set is empty or if \(A=B\). \(\square\)

The identity matters because the intersection is another measurable set, so \(\nu(A\cap B)\ge\alpha\). Rearranging the identity therefore yields an upper bound for the union:

In particular, if \(\nu(A)<\alpha+\varepsilon_A\) and \(\nu(B)<\alpha+\varepsilon_B\), where \(\varepsilon_A,\varepsilon_B>0\), then

The subtraction of \(\alpha\) is essential. Simply adding the two values would count the common part twice; the identity corrects for it, and the lower bound on the intersection controls that correction.

The Estimate for a Finite Union

Proof. We use induction on \(k\). For \(k=1\), the conclusion is exactly the assumed inequality. Suppose it holds for \(k-1\), where \(k\ge2\), and put \(F_{k-1}=\bigcup_{j=1}^{k-1}E_j\). The induction hypothesis gives \(\nu(F_{k-1})<\alpha+\sum_{j=1}^{k-1}\varepsilon_j\). The set \(F_{k-1}\cap E_k\) is measurable, so its value is at least \(\alpha\). Applying the union–intersection identity, \[ \nu(F_{k-1}\cup E_k) =\nu(F_{k-1})+\nu(E_k)-\nu(F_{k-1}\cap E_k) \le \nu(F_{k-1})+\nu(E_k)-\alpha. \] Using the two strict bounds on the right gives \(\nu(F_{k-1}\cup E_k)<\alpha+\sum_{j=1}^{k}\varepsilon_j\). Since \(F_{k-1}\cup E_k=\bigcup_{j=1}^k E_j\), this proves the induction step. The proof includes repeated or empty sets: each remains measurable, and the same lower bound on the intersection applies. \(\square\)

A useful choice is to allocate a total tolerance \(\varepsilon>0\) across the sets. If \(\varepsilon_j=\varepsilon/2^j\), then \(\sum_{j=1}^k\varepsilon_j<\varepsilon\), since the finite geometric sum is \(1-2^{-k}<1\). Consequently, any such finite union satisfies \(\nu(\bigcup_{j=1}^kE_j)<\alpha+\varepsilon\). The strict inequality in the selection of each set and the strict inequality in the finite geometric sum are both part of this conclusion.

A Pitfall: Signed Measures Are Not Monotone

For a positive measure, \(A\subseteq B\) implies \(\mu(A)\le\mu(B)\). That implication cannot be used for signed measures. For a concrete example, take \(X=\{a,b,c,d\}\) with counting sigma-algebra and atomic signed-measure weights \(-5,2,3,-2\), respectively. Then \(\{b,c,d\}\subseteq X\), but \(\nu(\{b,c,d\})=2+3-2=3\), whereas \(\nu(X)=-5+2+3-2=-2\). Monotonicity would require \(3\le-2\), which is false. Also, \(\varnothing\subseteq\{a\}\) while \(\nu(\varnothing)=0\) and \(\nu(\{a\})=-5\).

The correction is not to assume a direction of monotonicity. Instead, use the exact union–intersection identity and the global lower bound \(\nu(A\cap B)\ge\alpha\). This yields a valid upper bound for a union even when adding points can either raise or lower the signed measure. The estimate is useful precisely because it only needs a lower bound on values of measurable sets, not monotonicity.