A Uniform Lower Bound
A signed measure can assign negative values to measurable sets, so its values cannot be bounded below merely by appealing to nonnegativity. A useful starting point is a representation as a difference of two finite positive measures. The positive measure being subtracted has a finite total mass, and no measurable set can carry more than that mass. This gives a single lower bound that works for every measurable set.
Throughout the general discussion, let \((X,\mathcal{M})\) be a measurable space, with no assumption of completeness or sigma-finiteness. Let \(\alpha\) and \(\beta\) be finite positive measures on this space, meaning that \(\alpha(X)<\infty\) and \(\beta(X)<\infty\), and define the real-valued signed measure \(\mu\) by \(\mu(A)=\alpha(A)-\beta(A)\) for every \(A\in\mathcal{M}\). All statements about \(\mu(A)\) below hold for every measurable set \(A\), not merely almost everywhere.
The empty set belongs to \(\mathcal{M}\) and satisfies \(\mu(\varnothing)=0\). Thus \(S_\mu\) is nonempty. To show that its infimum is a real number, it remains to find a finite lower bound. The key estimate is \(\beta(A)\le\beta(X)\): since \(A\subseteq X\), the disjoint union \(X=A\cup A^c\) and positivity give \(\beta(X)=\beta(A)+\beta(A^c)\ge\beta(A)\).
The Lower-Bound Theorem
Proof. Fix any \(A\in\mathcal{M}\). Positivity gives \(\alpha(A)\ge0\) and \(\beta(A)\ge0\). Also, \(A\subseteq X\) implies \(\beta(A)\le\beta(X)\) and \(\alpha(A)\le\alpha(X)\). Therefore $$ \mu(A)=\alpha(A)-\beta(A)\ge 0-\beta(X)=-\beta(X) $$ and $$ \mu(A)=\alpha(A)-\beta(A)\le\alpha(X)-0=\alpha(X). $$ These are finite bounds because \(\alpha(X)\) and \(\beta(X)\) are finite. The set \(S_\mu\) is nonempty because \(0=\mu(\varnothing)\in S_\mu\). By the completeness property of the real numbers, every nonempty subset of \(\mathbb{R}\) bounded below has a real infimum. The lower bound just proved gives \(s_\mu\ge-\beta(X)\), and \(0\in S_\mu\) gives \(s_\mu\le0\). Hence \(s_\mu\) is finite and satisfies the asserted bounds. \(\square\)
The estimate applies uniformly: the same number \(-\beta(X)\) bounds \(\mu(A)\) for every \(A\), even though the set \(A\) may vary. It uses only positivity and finiteness of the two measures. No exchange of limits and integrals, and no assumption about a density or atoms, is involved.
Proof. Suppose \(\mu=\alpha_1-\beta_1=\alpha_2-\beta_2\), where all four measures are finite and positive on \((X,\mathcal{M})\). For every \(A\in\mathcal{M}\), both representations give the same real number \(\mu(A)\). Thus the sets \(\{\alpha_1(A)-\beta_1(A):A\in\mathcal{M}\}\) and \(\{\alpha_2(A)-\beta_2(A):A\in\mathcal{M}\}\) are equal. Equal subsets of \(\mathbb{R}\) have equal infima, so the infimum depends on \(\mu\), not on the chosen representation. \(\square\)
The lower bound \(-\beta(X)\), by contrast, does depend on the representation. It is always valid, but need not be the greatest lower bound. The following examples separate these two points.
Worked Example: Finding the Infimum on a Finite Space
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be the collection of all subsets of \(X\), and define positive measures by \(\alpha(\{a\})=3\), \(\beta(\{b\})=2\), \(\beta(\{c\})=5\), with all other singleton masses zero. Set \(\mu=\alpha-\beta\). Compute \(s_\mu\).
For a subset \(A\), its value is the sum of the masses of the points it contains: \(a\) contributes \(3\), \(b\) contributes \(-2\), and \(c\) contributes \(-5\). The eight subsets and their values are $$ \begin{aligned} \mu(\varnothing)&=0, & \mu(\{a\})&=3, & \mu(\{b\})&=-2, & \mu(\{c\})&=-5,\\ \mu(\{a,b\})&=3-2=1, & \mu(\{a,c\})&=3-5=-2, & \mu(\{b,c\})&=-2-5=-7, & \mu(X)&=3-2-5=-4. \end{aligned} $$ The smallest of these eight values is \(-7\), so \(s_\mu=-7\), attained at \(\{b,c\}\). Check: \(\alpha(X)=3\) and \(\beta(X)=2+5=7\), so the theorem gives \(-7\le\mu(A)\le3\) for every \(A\). The computed minimum is \(-7\), and \(\mu(\{b,c\})=0-(2+5)=-7\).
Worked Example: A Signed Measure Given by a Density
On \(X=[0,1]\) with the Lebesgue sigma-algebra and Lebesgue measure \(m\), define \(\mu(A)=\int_A(x-\tfrac12)\,dm\) for each measurable \(A\subseteq[0,1]\). Write \(\mu\) as a difference of finite positive measures and find its infimum over all measurable sets.
The density is nonnegative on \([\tfrac12,1]\) and nonpositive on \([0,\tfrac12]\). Define $$ \alpha(A)=\int_A(x-\tfrac12)^+\,dm,\qquad \beta(A)=\int_A(\tfrac12-x)^+\,dm, $$ where \(u^+=\max(u,0)\) for a real number \(u\). Both are positive measures. Their total masses are $$ \alpha(X)=\int_{1/2}^1(x-\tfrac12)\,dx =\left[\frac{x^2}{2}-\frac{x}{2}\right]_{1/2}^1 =0-\left(\frac18-\frac14\right)=\frac18, $$ and $$ \beta(X)=\int_0^{1/2}(\tfrac12-x)\,dx =\left[\frac{x}{2}-\frac{x^2}{2}\right]_0^{1/2} =\frac14-\frac18=\frac18. $$ Thus \(\mu=\alpha-\beta\), and the theorem gives \(\mu(A)\ge-\tfrac18\) for every measurable \(A\).
Take \(A=[0,\tfrac12]\). On this set the density is \(x-\tfrac12\), so $$ \mu(A)=\int_0^{1/2}(x-\tfrac12)\,dx =\left[\frac{x^2}{2}-\frac{x}{2}\right]_0^{1/2} =\frac18-\frac14=-\frac18. $$ Therefore \(s_\mu=-\tfrac18\). Check: on the full interval, \(\mu(X)=[x^2/2-x/2]_0^1=0\), which is consistent with \(\alpha(X)-\beta(X)=\tfrac18-\tfrac18=0\). The set used above has value exactly equal to the theorem's lower bound.
Worked Example: A Valid Bound That Is Not Sharp
Let \(X=\{0,1\}\) with all subsets measurable. Define \(\alpha=\delta_0+\delta_1\) and \(\beta=\delta_0+\delta_1\), where \(\delta_x(A)=1\) if \(x\in A\) and \(0\) otherwise. Then \(\mu=\alpha-\beta\) is the zero signed measure. Compare the theorem's bound with the actual infimum.
For each \(A\subseteq X\), \(\alpha(A)=\beta(A)\), so \(\mu(A)=0\). In particular, the values on \(\varnothing,\{0\},\{1\},X\) are \(0,0,0,0\), respectively, and \(s_\mu=0\). On the other hand, \(\beta(X)=\delta_0(X)+\delta_1(X)=1+1=2\), so the theorem's lower bound from this representation is \(-2\). It is correct because \(0\ge-2\), but it is not sharp. Check: \(\alpha(X)=2\) as well, so the full bound is \(-2\le\mu(A)\le2\); all four actual values lie within it.
Infimum, Bound, and Attainment
A lower bound and an infimum are related but are not interchangeable terms. The theorem constructs one lower bound directly from a chosen representation. The infimum is the greatest of all lower bounds for the set of values. In the finite-space example and the density example, the displayed lower bound is attained and therefore equals the infimum. In the zero-measure example, the representation gives a weaker bound than the actual infimum.
A common error is to treat \(-\beta(X)\) as the infimum merely because it is the lower bound supplied by the theorem. The zero-measure example gives a concrete countercalculation: substituting \(\beta(X)=2\) yields the valid estimate \(\mu(A)\ge-2\), but checking the four measurable sets gives \(\mu(A)=0\) every time. Hence the actual infimum is \(0\), not \(-2\). The correction is to use the representation to prove boundedness, and then determine the infimum from the full set of values only when an additional argument or computation is available.
For any representation \(\mu=\alpha-\beta\) as above, the conclusions established here are exactly $$ -\beta(X)\le \inf_{A\in\mathcal{M}}\mu(A)\le0. $$ The first inequality follows because every value is at least \(-\beta(X)\); the second follows because the empty set contributes the value \(0\). The representation-independent quantity is the infimum, while the particular numerical lower bound may change if a different pair of positive measures represents the same signed measure.
Check Your Understanding
In each question, work with the full collection of measurable sets indicated, and distinguish a lower bound from the actual infimum.
- Suppose \(\mu=\alpha-\beta\) on a measurable space and \(\alpha,\beta\) are finite positive measures. Derive the lower estimate for \(\mu(A)\) using \(A\subseteq X\).
- Why does \(\mu(\varnothing)=0\) imply that \(\inf_{A\in\mathcal{M}}\mu(A)\le0\)?
- On \(X=\{p,q\}\) with all subsets measurable, let \(\mu(\{p\})=4\) and \(\mu(\{q\})=-6\). List the four set values and find the infimum.
- If a representation gives \(\beta(X)=5\), what lower bound does the theorem provide? Does that information alone imply the infimum equals \(-5\)?
- Explain why two difference-of-positive-measures representations of the same set function must give the same infimum, even if their bounds \(-\beta(X)\) differ.