Signed Measure Definitions and Sign Conventions

What Makes a Measure Signed?

A positive measure assigns a nonnegative value to every measurable set. A signed measure retains countable additivity but allows negative values as well. That change matters: inclusion of sets no longer guarantees an inequality between their values. It also requires care with infinite values, since \(+\infty+(-\infty)\) is undefined.

In the general definition below, \((X,\mathcal{M})\) is a measurable space. A signed measure is defined on this space and does not require an ambient measure \(\mu\). When we use Lebesgue measure in examples, the space is \((\mathbb{R},\mathcal{L},m)\), where \(m\) is complete, \(\sigma\)-finite Lebesgue measure.

Definition: A signed measure on \((X,\mathcal{M})\) is a set function \(\nu:\mathcal{M}\to[-\infty,\infty]\) such that \(\nu(\varnothing)=0\), \(\nu\) is countably additive on disjoint measurable sets, and \(\nu\) does not take both the value \(+\infty\) and the value \(-\infty\). Countable additivity means that for pairwise disjoint \(E_1,E_2,\ldots\in\mathcal{M}\), the sum \(\sum_{n=1}^{\infty}\nu(E_n)\) is well-defined and \(\nu(\bigcup_{n=1}^{\infty}E_n)=\sum_{n=1}^{\infty}\nu(E_n)\).

For such a series, the positive and negative contributions must not both have infinite total magnitude. Equivalently, the sum is interpreted as the sum of its positive terms minus the sum of the magnitudes of its negative terms, and that expression cannot be \(\infty-\infty\). The restriction that \(\nu\) does not take both infinite values is part of the signed-measure convention; it prevents such an undefined subtraction at the level of set values as well.

If \(\nu(E)\in\mathbb{R}\) for every \(E\in\mathcal{M}\), we will say that \(\nu\) is finite-valued. This is different from an extended-real signed measure, which may take \(+\infty\) or may take \(-\infty\), but not both. Terminology such as “finite signed measure” is used with more than one convention, so the range or finiteness condition should be stated explicitly.

Definition: A positive measure on \((X,\mathcal{M})\) is a countably additive set function \(\mu:\mathcal{M}\to[0,\infty]\) with \(\mu(\varnothing)=0\). Thus every positive measure is also a signed measure under the definition above, but a signed measure need not be positive.

Finite Additivity and a Basic Construction

Countable additivity immediately gives finite additivity. The statement is valid even if one of the set values is infinite, because a signed measure cannot assign opposite infinite signs.

Proposition: Let \(\nu\) be a signed measure on \((X,\mathcal{M})\). If \(E,F\in\mathcal{M}\) are disjoint, then \(\nu(E\cup F)=\nu(E)+\nu(F)\). More generally, the same equality holds for any finite pairwise disjoint family.

Proof. Given disjoint \(E,F\in\mathcal{M}\), form the sequence \(E_1=E\), \(E_2=F\), and \(E_n=\varnothing\) for \(n\ge3\). Countable additivity gives \(\nu(E\cup F)=\sum_{n=1}^{\infty}\nu(E_n)=\nu(E)+\nu(F)\), since every later term is \(\nu(\varnothing)=0\). The finite-family statement follows by applying this two-set identity successively to the disjoint sets and their finite unions. No step adds \(+\infty\) to \(-\infty\), by the defining restriction on \(\nu\). \(\square\)

A useful way to build examples is to subtract two finite positive measures. The finiteness assumption ensures that the subtraction and the countable sums involved are unambiguous.

Proposition: Let \(\mu_1,\mu_2\) be finite positive measures on the same measurable space \((X,\mathcal{M})\). Then \(\nu(E)=\mu_1(E)-\mu_2(E)\) defines a finite-valued signed measure on \((X,\mathcal{M})\).

Proof. Since both measures are finite, \(\mu_1(E)\) and \(\mu_2(E)\) are real numbers for every \(E\in\mathcal{M}\); hence \(\nu(E)\in\mathbb{R}\). Also \(\nu(\varnothing)=0-0=0\). Let \((E_n)\) be pairwise disjoint measurable sets. Countable additivity of each positive measure gives \(\mu_i(\bigcup_n E_n)=\sum_n\mu_i(E_n)\) for \(i=1,2\). These two series have finite sums, since \(\sum_n\mu_i(E_n)=\mu_i(\bigcup_n E_n)\le\mu_i(X)<\infty\). Therefore their termwise difference is an absolutely convergent series: \(\sum_n|\mu_1(E_n)-\mu_2(E_n)|\le\sum_n\mu_1(E_n)+\sum_n\mu_2(E_n)<\infty\). It follows that \(\nu(\bigcup_nE_n)=\mu_1(\bigcup_nE_n)-\mu_2(\bigcup_nE_n)=\sum_n(\mu_1(E_n)-\mu_2(E_n))=\sum_n\nu(E_n)\). Thus \(\nu\) satisfies the signed-measure axioms. \(\square\)

Worked Example: A Signed Measure on the Lebesgue Space

On \((\mathbb{R},\mathcal{L},m)\), define \(\nu(E)=m(E\cap[0,1])-m(E\cap[2,3])\) for \(E\in\mathcal{L}\). Compute its values on the two intervals and their union, and test monotonicity.

The two set functions \(E\mapsto m(E\cap[0,1])\) and \(E\mapsto m(E\cap[2,3])\) are positive measures, each bounded above by \(1\), because each interval has Lebesgue measure \(1\). The proposition therefore shows that their difference is a finite-valued signed measure. Since \([0,1]\cap[2,3]=\varnothing\), we get \(\nu([0,1])=m([0,1])-m(\varnothing)=1-0=1\), and \(\nu([2,3])=m(\varnothing)-m([2,3])=0-1=-1\).

The union is disjoint, so \(\nu([0,1]\cup[2,3])=m([0,1])-m([2,3])=1-1=0\). To test monotonicity, note that \([0,1]\subset[0,1]\cup[2,3]\), but the values are \(1\) and \(0\), respectively. Thus the inclusion does not imply \(\nu([0,1])\le\nu([0,1]\cup[2,3])\): it would require \(1\le0\), which is false. Check: finite additivity gives \(1+(-1)=0\), exactly the value computed on the union.

Countable Additivity in a Concrete Partition

For a signed measure, countable additivity is an actual requirement, not merely a rule for finite unions. The next calculation checks an infinite disjoint union and shows how signed-measure additivity is applied without informal rearrangement of terms.

Worked Example: A Countable Partition of an Interval

Use the signed measure \(\nu\) just defined. For \(n\ge1\), let \(E_n=(1/(n+1),1/n]\), and let \(E_0=\{0\}\). Compute the series of set values and compare it with the value on the union.

The sets are pairwise disjoint. Their union is \([0,1]\): the intervals \(E_n\) cover every \(x\in(0,1]\), and \(E_0\) supplies the point \(0\). Each \(E_n\) lies in \([0,1]\) and does not meet \([2,3]\), so \(\nu(E_n)=m(E_n)=1/n-1/(n+1)=1/(n(n+1))\). Also \(\nu(E_0)=m(\{0\})=0\).

\sum_{n=0}^{\infty}\nu(E_n) =0+\sum_{n=1}^{\infty}\left(\frac1n-\frac1{n+1}\right) =\lim_{N\to\infty}\left(1-\frac1{N+1}\right) =1.

On the other hand, \(\nu(\bigcup_{n=0}^{\infty}E_n)=\nu([0,1])=m([0,1])-m(\varnothing)=1\). The values agree, as required by countable additivity. Check: the \(N\)-th partial sum is \(1-1/(N+1)\), which tends to \(1\); the omitted tail has value \(1/(N+1)\), tending to zero.

Infinite Values and the Prohibition on Opposite Signs

The range \([-\infty,\infty]\) in the definition does not mean that both infinite signs may occur. A signed measure may allow \(+\infty\), or it may allow \(-\infty\), but its countable-additivity formula must never require an undefined \(\infty-\infty\).

Worked Example: One Infinite Sign Is Permitted

On \((\mathbb{R},\mathcal{L},m)\), consider \(\nu(E)=m(E)\). This is a positive measure, hence also a signed measure, and it takes the value \(+\infty\). Verify countable additivity for a particular partition.

For \(k=0,1,2,\ldots\), let \(E_k=[k,k+1)\). These measurable sets are pairwise disjoint and their union is \([0,\infty)\). For every \(k\), the interval has length \(1\), so \(\nu(E_k)=m(E_k)=1\). Therefore \(\sum_{k=0}^{\infty}\nu(E_k)=\sum_{k=0}^{\infty}1=+\infty\). The union has infinite Lebesgue measure, so \(\nu([0,\infty))=+\infty\), in agreement with countable additivity. The set function \(E\mapsto-m(E)\) similarly permits \(-\infty\), but it never takes \(+\infty\). Check: the partial sum through \(k=N\) is \(N+1\), and its limit is \(+\infty\), matching the union's measure.

Worked Example: A Formula That Tries to Use Both Infinite Signs

On \((\mathbb{R},\mathcal{L},m)\), consider the tempting expression \(\eta(E)=m(E\cap[0,\infty))-m(E\cap(-\infty,0))\). Explain why this does not define a signed measure on all of \(\mathcal{L}\).

For \(E=[0,\infty)\), the expression gives \(+\infty-0=+\infty\), since \(m([0,\infty))=+\infty\) and \(m([0,\infty)\cap(-\infty,0))=0\). For \(E=(-\infty,0)\), it gives \(0-(+\infty)=-\infty\). Both sets are measurable, so the proposed set function takes both infinite signs, contrary to the definition. In particular, for \(E=\mathbb{R}\) the formula would require \(m([0,\infty))-m((-\infty,0))=+\infty-(+\infty)\), which is undefined. The expression therefore fails to assign a value to every measurable set and is not a signed measure. Check: the two half-lines are disjoint and have union \(\mathbb{R}\); adding their proposed values would require \(+\infty+(-\infty)\), not a defined extended-real sum.

Signed Values Are Not Positive-Measure Values

For a positive measure \(\mu\) on any measurable space \((X,\mathcal{M})\), inclusion \(E\subseteq F\) implies \(\mu(E)\le\mu(F)\): write \(F=E\cup(F\setminus E)\) as a disjoint union and use finite additivity and \(\mu(F\setminus E)\ge0\). The signed-measure identity still holds, but its second term can be negative, so this monotonicity argument does not apply.

The example on \((\mathbb{R},\mathcal{L},m)\) demonstrates the failure concretely: \([0,1]\subset[0,1]\cup[2,3]\), while \(\nu([0,1])=1\) and \(\nu([0,1]\cup[2,3])=0\). The calculation is consistent with finite additivity because the added set has value \(-1\). A signed measure may be negative on some measurable sets, but it must still satisfy its countable-additivity axiom with well-defined sums. Positivity and countable additivity are separate properties; one does not follow from the other.

Common pitfall. It is incorrect to infer that a signed measure is monotone just because it is additive. For the specific inclusion \([2,3]\subset[0,1]\cup[2,3]\), the values are \(-1\) and \(0\), and \(-1\le0\); this example does not disprove monotonicity. The correct test is \([0,1]\subset[0,1]\cup[2,3]\): its values are \(1\) and \(0\), and \(1\le0\) is false. Always compute both values before using an inclusion to claim that monotonicity fails.

Check Your Understanding

For each question, keep separate the axioms of countable additivity, the allowed range, and the extra nonnegativity condition for a positive measure.

State the range restriction that distinguishes a positive measure from a general signed measure.
Why is \(E\mapsto-m(E)\) on \((\mathbb{R},\mathcal{L},m)\) allowed to take \(-\infty\), even though it is not positive?
For the partition \(E_n=(1/(n+1),1/n]\), compute \(\nu(E_n)\) for the signed measure in the first worked example and find the sum over \(n\ge1\).
Give an inclusion in the first worked example for which the signed-measure values directly contradict monotonicity, and compute both values.
Explain why the expression \(m(E\cap[0,\infty))-m(E\cap(-\infty,0))\) fails to define a signed measure on all Lebesgue measurable sets.