Extensions from a Countable Partition

From Cell Weights to a Measure

A countable partition lets us build a measure by assigning a nonnegative weight to each cell and adding the weights. The main points are to specify which sets are being measured and to justify that the sum over infinitely many cells is countably additive. In particular, the weight belongs to an entire partition cell: if that cell contains several points, the construction does not automatically assign a measure to each of its subsets.

Let \(X\) be a set and let \((C_n)_{n\ge1}\) be a countably infinite partition of \(X\): each \(C_n\) is nonempty, the cells are pairwise disjoint, and \(X=\bigcup_{n\ge1}C_n\). Choose weights \(w_n\ge0\) such that \(W=\sum_{n=1}^{\infty}w_n<\infty\). The finite sum condition makes the total mass finite; it allows zero weights, but not negative weights.

Here an algebra is closed under finite unions and complements relative to \(X\), while a premeasure is a nonnegative, countably additive set function whenever a disjoint countable union and its union all belong to its domain. The sigma-algebra generated by the cells is larger than \(\mathcal{A}\): it allows unions of infinitely many cells. In this setting it has a particularly direct description.

Proof. Every union \(\bigcup_{n\in I}C_n\) is a countable union of sets \(C_n\), so it belongs to \(\sigma(\{C_n:n\ge1\})\). Conversely, the family of all such unions is a sigma-algebra: it contains \(X\) and \(\varnothing\); the complement of the union indexed by \(I\) is the union indexed by \(I^c=\mathbb{N}\setminus I\); and a countable union of sets indexed by \(I_1,I_2,\ldots\) is the union indexed by \(\bigcup_{j\ge1}I_j\). This sigma-algebra contains every \(C_n\), so it contains the sigma-algebra generated by the cells. The two inclusions prove the equality. \(\square\)

Write \(\mathcal{M}=\sigma(\{C_n:n\ge1\})\). Thus each \(E\in\mathcal{M}\) has a unique representation \(E=\bigcup_{n\in I_E}C_n\), with \(I_E\subseteq\mathbb{N}\). The proposed extension is the sum of the weights of all cells included in \(E\).

Proof. First, \(\mu(\varnothing)=0\), because its index set is empty. Let \((E_j)_{j\ge1}\) be pairwise disjoint sets in \(\mathcal{M}\), and write \(E_j=\bigcup_{n\in I_j}C_n\). The index sets \(I_j\) are pairwise disjoint: if an index \(n\) belonged to two of them, the nonempty cell \(C_n\) would be contained in both corresponding sets. Also, \(\bigcup_j E_j\) has index set \(\bigcup_j I_j\). Therefore \(\mu(\bigcup_jE_j)=\sum_{n\in\bigcup_jI_j}w_n=\sum_j\sum_{n\in I_j}w_n=\sum_j\mu(E_j)\). The equality of sums follows from the definition of a nonnegative series as the supremum of its finite subsums: finite collections of indices on the left lie in finitely many \(I_j\), and finite subsums from finitely many \(I_j\) are finite subsums on the left. This proves countable additivity. For \(X\), the index set is \(\mathbb{N}\), so \(\mu(X)=W<\infty\). Finally, if \(A\in\mathcal{A}\) has finite index set \(F\), the definition gives \(\mu(A)=\sum_{n\in F}w_n=\mu_0(A)\). Hence \(\mu\) is the claimed finite measure and extension. \(\square\)

The proof also verifies that \(\mu_0\) really is a premeasure. If a disjoint sequence of sets in \(\mathcal{A}\) has union \(A\in\mathcal{A}\), then \(A\) contains only finitely many cells. Each nonempty member of the sequence must contain at least one of those cells, and disjoint members cannot share a cell. Consequently only finitely many members are nonempty, and finite additivity gives the required countable-additivity identity. The case \(A=\varnothing\) has every member empty.

Computing Finite and Infinite Unions

Why This Is the Extension

The formula is not merely one possible assignment that happens to agree on finite unions. Since \(\mu_0(X)=W<\infty\), the premeasure is finite on the whole space. The finite-measure uniqueness theorem for extensions says that two measures on \(\sigma(\mathcal{A})\) agreeing with a finite premeasure on the generating algebra \(\mathcal{A}\) must agree on all of \(\sigma(\mathcal{A})\). This is the standard uniqueness result from the extension theory. Its hypotheses hold here: \(\mathcal{A}\) is an algebra, it generates \(\mathcal{M}\), and both the premeasure and the displayed extension have total mass \(W<\infty\). Thus any measure extension to \(\mathcal{M}\) has the cell-sum formula.

In particular, the measure space \((X,\mathcal{M},\mu)\) is finite, hence sigma-finite. It need not be complete. For example, in the interval partition above, every cell has positive weight, so the only measurable null set is \(\varnothing\); nevertheless, subsets of a nonempty cell need not be measurable. More generally, zero-weight cells can form measurable null sets, but subsets of those cells are not automatically in \(\mathcal{M}\). Completeness would require every subset of every measurable null set to be measurable, which is not guaranteed by this construction.

Common mistake. In the two-point-cell example, it is incorrect to reason that \(C_1\) has weight \(1/2\), so each of its two points has weight \(1/4\), and therefore \(\mu(\{(1,0)\})=1/4\). The construction assigns mass \(1/2\) to the whole cell \(C_1\), not separately to its points, and \(\{(1,0)\}\notin\mathcal{M}\). The correction is to first check that a proposed set belongs to the domain \(\mathcal{M}\); the formula applies only to unions of partition cells. If the cells are singletons, as in the first example, every subset is a union of cells and the formula does apply to every subset.

For any \(E\in\mathcal{M}\), the sum uses precisely the indices of cells contained in \(E\), each once. It is a finite sum when \(E\in\mathcal{A}\), and a nonnegative series for general \(E\). The condition \(\sum_nw_n<\infty\) guarantees \(\mu(E)\le W\) for every measurable \(E\); it does not require every individual weight to be positive.