Proving Agreement on the Original Ring

The Agreement to Be Proved

Let \(\mathcal{R}\) be a ring of subsets of a set \(X\): it contains \(\varnothing\) and is closed under finite unions and differences. Let \(\rho:\mathcal{R}\to[0,\infty]\) be a premeasure. The point of this tutorial is to prove that the covering construction does not change the value assigned to a set already in \(\mathcal{R}\). The essential step is to turn any countable cover of that set into a disjoint sequence of ring sets to which premeasure additivity applies.

Definition: A premeasure on \(\mathcal{R}\) is a set function \(\rho:\mathcal{R}\to[0,\infty]\) with \(\rho(\varnothing)=0\), such that whenever \(E_1,E_2,\ldots\) are pairwise disjoint members of \(\mathcal{R}\) and \(E=\bigcup_{n=1}^{\infty}E_n\) also belongs to \(\mathcal{R}\), then \(\rho(E)=\sum_{n=1}^{\infty}\rho(E_n)\). The sum is allowed to be \(+\infty\).

For \(A\subseteq X\), a countable sequence \((A_n)\) of members of \(\mathcal{R}\) is an admissible cover of \(A\) if \(A\subseteq\bigcup_{n=1}^{\infty}A_n\). The induced covering outer measure is

\rho^*(A)=\inf\left\{\sum_{n=1}^{\infty}\rho(A_n): A_n\in\mathcal{R},\quad A\subseteq\bigcup_{n=1}^{\infty}A_n\right\}.

An admissible cover need not exist for every subset \(A\subseteq X\). If none exists, the family whose infimum is taken is empty, and we use the convention that its infimum is \(+\infty\). For a set \(E\in\mathcal{R}\), however, there is always an admissible cover: take \(A_1=E\) and \(A_n=\varnothing\) for \(n\ge2\). No assumption that \(\rho\) is finite or sigma-finite is needed for the agreement result.

Two Facts About the Premeasure

Before considering a cover, we need monotonicity on the ring. It follows directly from the definition of premeasure and does not require a separate assumption.

Proposition: If \(B,A\in\mathcal{R}\) and \(B\subseteq A\), then \(\rho(B)\le\rho(A)\).

Proof. Since \(\mathcal{R}\) is a ring, \(A\setminus B\in\mathcal{R}\). The two sets \(B\) and \(A\setminus B\) are disjoint and their union is \(A\in\mathcal{R}\). Premeasure additivity gives \(\rho(A)=\rho(B)+\rho(A\setminus B)\). The second term is nonnegative, so \(\rho(B)\le\rho(A)\), also when one or both values are infinite. \(\square\)

The other fact we will use is that a finite union of ring sets and the difference of two ring sets are again in the ring. In particular, intersections are in the ring, since \(C\cap D=C\setminus(C\setminus D)\) for \(C,D\in\mathcal{R}\). These closure properties ensure that the disjoint pieces constructed from a cover are eligible inputs to \(\rho\).

Agreement on Every Ring Set

Theorem: For every \(E\in\mathcal{R}\), the covering outer measure induced by \(\rho\) satisfies \(\rho^*(E)=\rho(E)\).

Proof. First use the admissible cover \(A_1=E\), \(A_n=\varnothing\) for \(n\ge2\). Its total cost is \(\rho(E)+\sum_{n\ge2}\rho(\varnothing)=\rho(E)\), so the infimum satisfies \(\rho^*(E)\le\rho(E)\).

For the reverse inequality, fix any admissible cover \((A_n)\) of \(E\), so each \(A_n\in\mathcal{R}\) and \(E\subseteq\bigcup_{n=1}^{\infty}A_n\). Define, using the union of all earlier covering sets,

B_n=E\cap\left(A_n\setminus\bigcup_{j=1}^{n-1}A_j\right),

where the union over no indices for \(n=1\) is \(\varnothing\). Each \(B_n\) belongs to \(\mathcal{R}\): the finite union of earlier \(A_j\)'s is in the ring, its difference from \(A_n\) is in the ring, and intersection with \(E\in\mathcal{R}\) stays in the ring. The \(B_n\) are pairwise disjoint. Indeed, if \(i<n\), then \(B_i\subseteq A_i\), whereas \(B_n\) excludes every point in \(A_i\).

They also have union exactly \(E\). Each \(B_n\subseteq E\), so their union is contained in \(E\). Conversely, for any \(x\in E\), admissibility gives at least one index \(n\) with \(x\in A_n\). Choose the least such index. Then \(x\notin A_j\) for \(j<n\), so \(x\in B_n\). Thus \(E=\bigcup_{n=1}^{\infty}B_n\). Since \(E\in\mathcal{R}\), premeasure additivity applies and yields \(\rho(E)=\sum_{n=1}^{\infty}\rho(B_n)\).

For each \(n\), \(B_n\subseteq A_n\), and both sets are in \(\mathcal{R}\). Monotonicity therefore gives \(\rho(B_n)\le\rho(A_n)\). Summing these termwise inequalities for nonnegative extended-real terms gives

\rho(E)=\sum_{n=1}^{\infty}\rho(B_n) \le \sum_{n=1}^{\infty}\rho(A_n).

This holds for every admissible cover, whether or not any cover attains the infimum. Therefore \(\rho(E)\) is a lower bound for all cover costs, and hence \(\rho(E)\le\rho^*(E)\). Combined with the first inequality, this proves \(\rho^*(E)=\rho(E)\). \(\square\)

The argument does not claim that \(\rho^*(A)\) is finite, or even that a cover exists, for every \(A\subseteq X\). It proves equality only for \(E\in\mathcal{R}\), where the explicit one-set cover is available. It also proves only agreement of values on the ring; whether its sets satisfy Caratheodory's measurability criterion is a separate question.

Worked Applications

Worked Example: A Weighted Ring on Three Points

Take \(X=\{a,b,c\}\), \(\mathcal{R}=\mathcal{P}(X)\), and assign weights \(w(a)=2\), \(w(b)=1\), and \(w(c)=4\). Define \(\rho(S)=\sum_{x\in S}w(x)\). This is a premeasure: for disjoint sets whose union is in \(\mathcal{R}\), each point in the union occurs in exactly one set, so adding their weights gives the weight of the union.

Let \(E=\{a,c\}\). Its premeasure is \(\rho(E)=2+4=6\), and the single-set cover gives \(\rho^*(E)\le6\). To see the cover argument with overlap, use \(A_1=\{a,b\}\), \(A_2=\{b,c\}\), and \(A_n=\varnothing\) for \(n\ge3\). Their costs are \(\rho(A_1)=2+1=3\), \(\rho(A_2)=1+4=5\), and zero thereafter, for a total of \(8\). The disjointified pieces of \(E\) are \(B_1=\{a\}\) and \(B_2=\{c\}\), with costs \(2\) and \(4\); thus their sum is \(6\le8\). The theorem says every admissible cover costs at least \(6\), so \(\rho^*(E)=6\). Check: the explicit cover of \(E\) costs \(2+4=6\), while the overlapping cover costs \(3+5=8\); the infimum is bounded above by \(6\) and below by \(6\).

Worked Example: Counting Measure on Finite Subsets

Let \(X=\mathbb{N}\), let \(\mathcal{R}\) consist of all finite subsets of \(\mathbb{N}\), and set \(\rho(A)=|A|\). This is a premeasure on \(\mathcal{R}\): if pairwise disjoint finite sets have a finite union, only finitely many can be nonempty, and the cardinality of that union is the sum of their cardinalities.

Set \(E=\{2,5,9\}\). Then \(\rho(E)=3\). Consider the admissible cover \(A_1=\{2,5\}\), \(A_2=\{5,7\}\), \(A_3=\{9\}\), and \(A_n=\varnothing\) for \(n\ge4\). Its total cost is \(2+2+1=5\). The disjointified pieces relative to \(E\) are \(B_1=\{2,5\}\), \(B_2=\varnothing\), \(B_3=\{9\}\), and empty sets thereafter. Their total cardinality is \(2+0+1=3\), the cardinality of \(E\). The theorem gives \(\rho^*(E)\ge3\), and the cover \(E\) itself gives \(\rho^*(E)\le3\). Hence \(\rho^*(E)=3\). Check: the three distinct points of \(E\) each contribute one to \(\rho(E)\); the repeated point \(5\) in the displayed cover does not create an additional point in its disjointification.

Worked Example: A Subset with No Admissible Cover

Let \(X=\{1,2\}\), \(\mathcal{R}=\{\varnothing,\{1\}\}\), and define \(\rho(\varnothing)=0\), \(\rho(\{1\})=1\). This is a ring. It is a premeasure: a disjoint sequence in \(\mathcal{R}\) has at most one nonempty term, so if its union is in \(\mathcal{R}\), the required additivity holds.

For \(A=\{2\}\), no admissible cover exists. Every member of \(\mathcal{R}\) is a subset of \(\{1\}\), so any countable union of such members is also a subset of \(\{1\}\) and cannot contain \(2\). The infimum is therefore over the empty family, and by convention \(\rho^*(\{2\})=+\infty\). This does not conflict with the agreement theorem, because \(\{2\}\notin\mathcal{R}\). Check: the only possible covering sets are \(\varnothing\) and \(\{1\}\); neither, nor any countable union of them, contains \(2\).

A Common Disjointification Error

A frequent mistake is to subtract only the immediately preceding cover set, rather than the union of all earlier sets. In the weighted example above, consider the cover \(A_1=\{a,b\}\), \(A_2=\{b,c\}\), \(A_3=\{a,c\}\) of \(E=\{a,b,c\}\). Its total cost is \(3+5+6=14\), and \(\rho(E)=2+1+4=7\).

The incorrect pieces \(D_1=A_1\), \(D_2=A_2\setminus A_1=\{c\}\), and \(D_3=A_3\setminus A_2=\{a\}\) are not disjoint: \(a\in D_1\cap D_3\). Their costs sum to \(3+4+2=9\), not \(7\). The correction is to remove all earlier sets: \(B_n=E\cap(A_n\setminus\bigcup_{j<n}A_j)\). This gives \(B_1=\{a,b\}\), \(B_2=\{c\}\), and \(B_3=\varnothing\), which are disjoint and have total cost \(3+4+0=7=\rho(E)\). The disjointness is what permits premeasure additivity; without it, the equality for the pieces is unavailable.

Why Agreement Matters

The covering infimum is constructed using arbitrary countable covers, while the premeasure is initially defined only on \(\mathcal{R}\). Agreement shows that the construction preserves every original value, rather than silently changing the set function on its starting domain. The proof's lower bound is the substantive part: it applies to every cover, including covers with overlaps and covers whose costs approach an infimum without attaining it. Its key requirements are exactly the ring closure needed to form the pieces and countable additivity of the premeasure when their union is a ring set.

Check Your Understanding

For each question, identify which property of the ring or premeasure is being used.

Why does \(B\subseteq A\), with \(A,B\in\mathcal{R}\), imply \(\rho(B)\le\rho(A)\)?
For an admissible cover of \(E\in\mathcal{R}\), explain why the least index \(n\) with \(x\in A_n\) places \(x\) in \(B_n\).
Why is it necessary to subtract the union of all earlier cover sets in the definition of \(B_n\)?
In the finite-subset example, compute the total cost of the displayed cover and the total cardinality of its disjointified pieces.
Can a subset \(A\subseteq X\) have \(\rho^*(A)=+\infty\) because there is no admissible cover? Explain why this does not contradict agreement on \(\mathcal{R}\).