Finite Atomic Premeasures from Weights

Weights on a Finite Partition

A finite partition turns a measure-extension problem into a finite list of weights. The algebra generated by the partition consists of unions of whole cells, so a nonnegative weight on each cell specifies the value of a set function on every member of that algebra. The outer-measure construction then reveals an important distinction: a subset may split a cell and still be measurable if that cell has weight zero.

Let \(X\) be a set and let \(P_1,\ldots,P_N\) be a finite partition of \(X\): each \(P_i\) is nonempty, the cells are pairwise disjoint, and their union is \(X\). Let \(w_1,\ldots,w_N\) be finite nonnegative real numbers. Write \(\mathcal{A}\) for the algebra of all unions of these cells, including \(\varnothing\) and \(X\). For \(A\in\mathcal{A}\), define its weight by summing the weights of the cells it contains.

The sum over no indices is zero. The set function \(\mu_0\) is a finite premeasure on \(\mathcal{A}\), and its total mass is \(W=\sum_{i=1}^N w_i\). Here “premeasure” means that \(\mu_0(\varnothing)=0\) and that \(\mu_0\) is countably additive whenever a countable disjoint union of sets in \(\mathcal{A}\) is itself in \(\mathcal{A}\).

Proof. The empty set contains no cells, so \(\mu_0(\varnothing)=0\). Let \((A_k)_{k\ge1}\) be pairwise disjoint members of \(\mathcal{A}\), and suppose their union \(A\) belongs to \(\mathcal{A}\). Each nonempty \(A_k\) contains at least one cell. No cell can be contained in two different \(A_k\), since the sets are disjoint. There are only \(N\) cells, so at most \(N\) of the \(A_k\) are nonempty. Each cell contained in \(A\) is contained in exactly one of those nonempty sets. Consequently, summing the cell weights in the union counts each such weight exactly once, and \(\mu_0(A)=\sum_{k\ge1}\mu_0(A_k)\). Finally, \(X\) contains every cell, so \(\mu_0(X)=\sum_{i=1}^Nw_i=W<\infty\). \(\square\)

The Outer Measure and Its Measurable Sets

The covering construction starts with sets in \(\mathcal{A}\), even though the set to be covered can be any subset of \(X\). Define the associated outer measure \(\mu_0^*\) on the power set of \(X\) by taking the infimum of the total costs of countable algebra covers.

Proof. First fix \(E\subseteq X\). For an upper bound, cover \(E\) by the union of all cells it meets. This is a member of \(\mathcal{A}\), and its \(\mu_0\)-weight is \(\sum_{\{i:\,E\cap P_i\ne\varnothing\}}w_i\). Therefore \(\mu_0^*(E)\) is at most this sum.

For the reverse bound, consider any countable cover \(E\subseteq\bigcup_k A_k\) with \(A_k\in\mathcal{A}\). If \(E\) meets \(P_i\), choose \(x\in E\cap P_i\). Some \(A_k\) contains \(x\). Since \(A_k\) is a union of whole cells, it contains \(P_i\), and its weight includes \(w_i\). Summing the costs of the cover therefore counts each weight for a cell met by \(E\) at least once. All weights are nonnegative, so \(\sum_k\mu_0(A_k)\ge\sum_{\{i:\,E\cap P_i\ne\varnothing\}}w_i\). Taking the infimum proves the formula.

By definition, \(E\) is Carathéodory-measurable when \(\mu_0^*(T)=\mu_0^*(T\cap E)+\mu_0^*(T\setminus E)\) for every \(T\subseteq X\). Suppose first that \(E\) splits a cell \(P_i\) with \(w_i>0\), meaning both \(E\cap P_i\) and \(P_i\setminus E\) are nonempty. Take \(T=P_i\). The formula just proved gives \(\mu_0^*(T)=w_i\), while each of \(T\cap E\) and \(T\setminus E\) meets \(P_i\), so each has outer measure \(w_i\). Their sum is \(2w_i>w_i\), and the measurability condition fails.

Conversely, suppose no positive-weight cell is split. Fix any \(T\subseteq X\). For each positive-weight cell met by \(T\), the whole cell lies either in \(E\) or in \(X\setminus E\). It therefore contributes its weight to exactly one of \(\mu_0^*(T\cap E)\) and \(\mu_0^*(T\setminus E)\). Cells of weight zero contribute zero to all three outer measures, whether or not they are split. Using the formula for outer measure gives the Carathéodory equality for this \(T\). Since \(T\) was arbitrary, \(E\) is measurable. This proves both directions. \(\square\)

Carathéodory’s theorem, established earlier in the course, says that the Carathéodory-measurable sets form a sigma-algebra and that the outer measure restricted to them is a measure. Denote this measure space by \((X,\mathcal{M},\mu)\), where \(\mathcal{M}\) is that sigma-algebra and \(\mu(E)=\mu_0^*(E)\) for \(E\in\mathcal{M}\). Every member of \(\mathcal{A}\) is measurable, and the outer-measure formula shows \(\mu(A)=\mu_0(A)\), so this is an extension of the original premeasure. The measure is finite because \(\mu(X)=W\). If every weight is zero, the criterion says every subset of \(X\) is measurable and the extension assigns measure zero to every subset.

Indicators and Simple-Function Integrals

For a measurable set \(E\in\mathcal{M}\), its indicator \(\chi_E\) is the function equal to \(1\) on \(E\) and \(0\) on \(X\setminus E\). It is a measurable, real-valued simple function, and its integral is \(\int\chi_E\,d\mu=\mu(E)\). A simple function is a measurable real-valued function taking only finitely many values. When written as \(\varphi=\sum_{j=1}^r c_j\chi_{E_j}\) for pairwise disjoint measurable sets \(E_j\) and real coefficients \(c_j\), its Lebesgue integral is \(\int\varphi\,d\mu=\sum_{j=1}^r c_j\mu(E_j)\). These integrals are finite here because \(\mu(X)=W<\infty\) and the function has finitely many finite values.

Proof. Fix \(i\) with \(w_i>0\). If \(\varphi\) took two different values at points \(x,y\in P_i\), there would be a real number \(t\) strictly between those values. The measurable set \(\{z:\varphi(z)>t\}\) would contain one of \(x,y\) and not the other, splitting \(P_i\). This contradicts the measurability criterion. Thus \(\varphi\) is constant on \(P_i\). On a zero-weight cell, the function may vary, since any subset of that cell is measurable.

Let \(c_i\) be the constant value on each positive-weight cell. Each level set of \(\varphi\) is measurable. In the simple-function integral formula, its measure is the sum of the weights of the positive cells on which that level is attained; portions of zero-weight cells add no weight. Grouping the finitely many terms by positive-weight cell gives \(\int\varphi\,d\mu=\sum_{\{i:\,w_i>0\}}w_ic_i\). Since \(c_i=\varphi(x_i)\) for every \(x_i\in P_i\), the stated formula follows. \(\square\)

A Common Pitfall

A split cell does not automatically make a set nonmeasurable: the weight of the cell matters. For the set \(D=\{a,b,c\}\) above, it would be wrong to conclude “\(D\) splits \(P_2\), so \(D\) is not measurable.” The Carathéodory test shows why. The correct complement is \(D^c=\{d,e\}\), not a set that also contains \(b\). The outer-measure calculations are \(\mu_0^*(D)=2+0=2\) and \(\mu_0^*(D^c)=0+5=5\), whose sum is \(7=\mu_0^*(X)\). More generally, splitting a positive-weight cell fails the test by the strict inequality \(2w_i>w_i\); splitting a zero-weight cell contributes \(0\) on both sides and does not cause failure.

This finite model separates the roles of the algebra, the premeasure, and the extension. The premeasure records the assigned weights on unions of whole cells. Its covering outer measure depends only on which cells a set meets. Carathéodory measurability then allows arbitrary variation inside zero-weight cells, and integration reduces to a finite weighted sum over positive-weight cells. This is useful whenever a finite discrete model is used to calculate measures or expectations: zero-weight points or cells can be disregarded in the integral, but their measurability still follows from the extension.