Extracting a Negative Set at the Infimum

The Infimum as a Route to a Negative Set

For a signed measure, a set with small measure need not be negative: it may contain parts with both positive and negative contributions. A useful construction begins instead with the infimum of all set values. Sets whose values approach that infimum can be combined so that their limiting construction attains it. The minimizing property then forces every measurable subset of the resulting set to have nonpositive measure.

Throughout this tutorial, \((X,\mathcal{M})\) is a measurable space and \(\nu\) is a finite signed measure: \(\nu\) is countably additive and real-valued on \(\mathcal{M}\). By the lower-bound result for finite signed measures, the values \(\nu(E)\), \(E\in\mathcal{M}\), have a finite uniform lower bound. Define \(\alpha=\inf\{\nu(E):E\in\mathcal{M}\}\). Since \(\nu(\varnothing)=0\), \(\alpha\leq 0\), and the lower bound ensures \(\alpha\in\mathbb{R}\). All sets and set inclusions below are measurable, and all measure equalities are exact, not almost-everywhere statements.

Definition: A measurable set \(N\in\mathcal{M}\) is a negative set for \(\nu\) if \(\nu(B)\leq 0\) for every measurable \(B\subseteq N\). A set \(A\in\mathcal{M}\) attains the infimum if \(\nu(A)=\alpha\).

We use two continuity properties of finite signed measures. If \(E_n\uparrow E\), meaning \(E_n\subseteq E_{n+1}\) and \(E=\bigcup_nE_n\), then \(\nu(E_n)\to\nu(E)\): write the union as the disjoint union of \(E_1\) and \(E_k\setminus E_{k-1}\), \(k\geq2\), and apply countable additivity to the partial sums. If \(E_n\downarrow E\) and \(\nu(E_1)\) is finite, then \(\nu(E_n)\to\nu(E)\), by applying continuity from below to \(E_n^c\uparrow E^c\) and using \(\nu(E_n)=\nu(X)-\nu(E_n^c)\). These properties are used only for the real-valued finite signed measure specified above.

Building the Set from Near-Minimizers

For any \(\varepsilon>0\), the definition of \(\alpha\) gives a measurable set \(E\) such that \(\nu(E)<\alpha+\varepsilon\). Indeed, if no such set existed, \(\alpha+\varepsilon\) would be a lower bound larger than the infimum. Choose positive numbers \(\varepsilon_n\) with \(\sum_{n=1}^{\infty}\varepsilon_n<\infty\), and choose \(E_n\in\mathcal{M}\) so that \(\nu(E_n)<\alpha+\varepsilon_n\).

The finite-union estimate from finite unions of nearly minimizing sets says that for any finite collection \(E_r,\ldots,E_s\) chosen this way,

\alpha\leq \nu\!\left(\bigcup_{k=r}^{s}E_k\right) \leq \alpha+\sum_{k=r}^{s}\varepsilon_k.

The lower bound is the definition of \(\alpha\). For the upper bound, the one-step identity \(\nu(F\cup E)=\nu(F)+\nu(E)-\nu(F\cap E)\) gives \(\nu(F\cup E)\leq\nu(F)+\nu(E)-\alpha\), since \(\nu(F\cap E)\geq\alpha\). Applying this inequality successively proves the displayed finite-union bound.

Theorem: Let \(\nu\) be a finite signed measure on \((X,\mathcal{M})\), and let \(\alpha=\inf_{E\in\mathcal{M}}\nu(E)\). There exists a measurable negative set \(A\) such that \(\nu(A)=\alpha\).

Proof. Choose summable errors \(\varepsilon_n>0\) and sets \(E_n\in\mathcal{M}\) with \(\nu(E_n)<\alpha+\varepsilon_n\). For integers \(m\geq n\), put \(U_{n,m}=\bigcup_{k=n}^{m}E_k\), and put \(T_n=\bigcup_{k=n}^{\infty}E_k\). The sets \(U_{n,m}\) increase to \(T_n\) as \(m\to\infty\). By the finite-union estimate and continuity from below,

\alpha\leq \nu(T_n) =\lim_{m\to\infty}\nu(U_{n,m}) \leq \alpha+\sum_{k=n}^{\infty}\varepsilon_k.

The last inequality follows because every finite tail sum is at most the full tail sum; taking a limit preserves the non-strict upper bound. Now let \(A=\bigcap_{n=1}^{\infty}T_n\). The sets \(T_n\) decrease, and \(\nu(T_1)\) is finite, so continuity from above applies. Since the tails of a convergent nonnegative series tend to zero, the bounds just obtained imply \(\nu(T_n)\to\alpha\). Consequently,

\nu(A)=\lim_{n\to\infty}\nu(T_n)=\alpha.

It remains to prove that \(A\) is negative. Let \(B\in\mathcal{M}\) satisfy \(B\subseteq A\). The disjoint union \(A=B\cup(A\setminus B)\) gives \(\nu(A)=\nu(B)+\nu(A\setminus B)\). By the definition of \(\alpha\), \(\nu(A\setminus B)\geq\alpha=\nu(A)\). Subtracting \(\nu(A)\) from both sides of the additivity identity yields \(\nu(B)\leq0\). This also covers \(B=\varnothing\). Thus every measurable subset of \(A\) has nonpositive measure, as required. \(\square\)

Worked Constructions

Worked Example: Alternating Positive Atoms and a Negative Core

Let \(X=\{1^-,2^-,\ldots\}\cup\{1^+,2^+,\ldots\}\), with \(\mathcal{M}\) the power set. Give \(k^-\) mass \(-2^{-k}\) and \(k^+\) mass \(2^{-k}\), and define \(\nu(E)\) as the sum of the masses of the points in \(E\). This series is absolutely convergent because \(\sum_{k\geq1}2^{-k}=1\); thus \(\nu\) is a finite signed measure.

Let \(N=\{k^-:k\geq1\}\). Any set has negative contribution at least \(-1\), and its positive contribution is nonnegative. Therefore \(\nu(E)\geq-1\) for every \(E\), while \(\nu(N)=-\sum_{k\geq1}2^{-k}=-1\). Hence \(\alpha=-1\). Choose \(E_n=N\cup\{n^+\}\). Then \(\nu(E_n)=-1+2^{-n}\). With \(\varepsilon_n=2^{1-n}\), the required strict approximation holds because \(2^{-n}<2^{1-n}\), and \(\sum_{n\geq1}\varepsilon_n=2\).

The tail union is \(T_n=N\cup\{k^+:k\geq n\}\), so \(\nu(T_n)=-1+\sum_{k=n}^{\infty}2^{-k}=-1+2^{1-n}\). Thus \(T_n\downarrow N\), and the intersection \(A\) has \(\nu(A)=-1=\alpha\). Every measurable subset of \(A=N\) has measure a sum of terms \(-2^{-k}\), so it has measure at most zero. Check: the ordinary union \(\bigcup_nE_n\) is all of \(X\) and has measure \(-1+1=0\), not \(-1\). The tail-union intersection, not the ordinary union, is the minimizing set.

Worked Example: A Signed Measure Defined by a Density

On \(([0,1],\mathcal{L})\) with Lebesgue measure \(m\), define \(\nu(E)=\int_E(2x-1)\,dm\). The density is integrable and real-valued. For any measurable \(E\), the contribution from \([0,\tfrac12)\) is at least the integral over the whole negative region, and the contribution from \([\tfrac12,1]\) is nonnegative. Therefore \(\nu(E)\geq\int_0^{1/2}(2x-1)\,dx=-\tfrac14\). The set \([0,\tfrac12)\) attains this bound, so \(\alpha=-\tfrac14\).

For \(n\geq1\), set \(a_n=\tfrac12-\tfrac1{n+2}\) and \(E_n=[0,a_n]\). Direct integration gives

\nu(E_n)=\int_0^{a_n}(2x-1)\,dx =a_n^2-a_n =-\frac14+\frac1{(n+2)^2}.

Take \(\varepsilon_n=2/(n+2)^2\). The error is strictly smaller than \(\varepsilon_n\), and the errors are summable: since \(1/(n+2)^2\leq1/((n+1)(n+2))\),

\sum_{n=1}^{\infty}\varepsilon_n \leq 2\sum_{n=1}^{\infty}\left(\frac1{n+1}-\frac1{n+2}\right)=1.

The endpoints \(a_n\) increase to \(\tfrac12\). For each \(n\), the union of the sets \(E_k\) over \(k\geq n\) is \([0,\tfrac12)\): every point \(x<\tfrac12\) is eventually below \(a_k\), and no \(a_k\) reaches \(\tfrac12\). Hence the intersection of these tail unions is \(A=[0,\tfrac12)\). Its measure is \(\nu(A)=-\tfrac14\), and for every measurable \(B\subseteq A\), \(2x-1\leq0\) on \(B\), so \(\nu(B)\leq0\). Check: integrating gives \(\int_0^{1/2}(2x-1)\,dx=[x^2-x]_0^{1/2}=\tfrac14-\tfrac12=-\tfrac14\), the claimed infimum.

Worked Example: An Infimum Already Attained on a Finite Space

Let \(X=\{a,b,c\}\) with its power-set sigma-algebra, and assign masses \(\nu(\{a\})=-2\), \(\nu(\{b\})=1\), and \(\nu(\{c\})=-1\). For any subset, omitting a negative-mass point can only increase its value, and including \(b\) adds \(1\). Thus the smallest possible value is \(-2-1=-3\), attained by \(A=\{a,c\}\).

Choose \(E_n=A\) and \(\varepsilon_n=2^{-n}\). Then \(\nu(E_n)=-3<-3+2^{-n}=\alpha+\varepsilon_n\), and the error series sums to \(1\). Every tail union and their intersection equal \(A\), so the construction returns the minimizing set. To check negativity directly, the measurable subsets of \(A\) are \(\varnothing,\{a\},\{c\},\{a,c\}\), with values \(0,-2,-1,-3\), respectively; all are nonpositive.

A Common Pitfall: Taking the Wrong Limit of Sets

It is not enough to take the ordinary union of sets whose values approach the infimum. In the atomic example, every \(E_n\) has value \(-1+2^{-n}\), but their ordinary union has value \(0\). The proof instead forms each tail union \(T_n=\bigcup_{k\geq n}E_k\), then takes the decreasing intersection \(\bigcap_nT_n\). A point belongs to this intersection exactly when it belongs to infinitely many \(E_k\). The two set operations have different membership conditions, and the distinction can change the measure.

A second point of care is the strict approximation. The infimum property gives \(\nu(E_n)<\alpha+\varepsilon_n\), not an automatically strict inequality after taking a limit of finite-union estimates. The proof only needs the non-strict bound \(\nu(T_n)\leq\alpha+\sum_{k=n}^{\infty}\varepsilon_k\), which follows by passing to a limit. Summability makes this upper error tend to zero; merely choosing errors that tend to zero would not control the finite-union estimates.

The construction produces one negative set attaining the infimum without assuming a density or a countable underlying space. Its key ingredients are the lower bound defining \(\alpha\), summable approximation errors, and continuity of the finite signed measure along increasing and decreasing sequences. The next step in the theory uses such a negative set to divide the whole space into positive and negative parts.

Check Your Understanding

In each question, distinguish the infimum property from the additional argument that establishes negativity.

Why does \(\alpha=\inf_{E\in\mathcal{M}}\nu(E)\) satisfy \(\alpha\leq0\), and which assumption ensures that \(\alpha\) is finite?
For the atomic example, compute the value of \(T_n\) and explain why \(\bigcap_nT_n=N\), while \(\bigcup_nE_n=X\).
Show that if \(A\) attains the infimum, then every measurable \(B\subseteq A\) has \(\nu(B)\leq0\).
In the density example, verify that \(1/(n+2)^2\leq1/((n+1)(n+2))\) and use the telescoping sum to bound the total approximation error.
Why does continuity from above apply to \(T_n\) in the proof, and where is finiteness used?