Decomposition by Singleton Masses
On a countable space, a measure is determined by its values at individual points. This makes the Lebesgue decomposition relative to a reference measure especially concrete: points with positive reference mass support the absolutely continuous part, while points of zero reference mass can support only the singular part. The key is to separate these two sets before attempting to compute a density.
Throughout, let \(X\) be a finite or countably infinite set, equipped with the sigma-algebra \(\mathcal{P}(X)\) of all its subsets. Let \(\mu\) and \(\nu\) be positive, sigma-finite measures on \((X,\mathcal{P}(X))\). They need not be finite. Sigma-finiteness ensures that each singleton has finite measure: if \(X\) is covered by measurable sets of finite measure, each point belongs to one of those sets, so its singleton has finite measure too. This convention lets us use ordinary finite ratios at points with positive reference mass.
For each \(A\subseteq X\), countable additivity gives \(\mu(A)=\sum_{x\in A}\mu(\{x\})\), and likewise \(\nu(A)=\sum_{x\in A}\nu(\{x\})\). All such sums have nonnegative terms and are allowed to equal \(+\infty\). In particular, \(\mu(N)=0\), because every singleton in \(N\) has \(\mu\)-measure zero. The set \(S\) is allowed to be empty, as is \(N\).
The Pointwise Decomposition Formula
Proof. The two formulas defining \(\nu_{\mathrm{ac}}\) and \(\nu_{\mathrm{s}}\) give measures: if \(A_1,A_2,\ldots\) are pairwise disjoint subsets of \(X\), their union divides the singleton terms in either defining sum into disjoint groups, so summing over the union equals the sum of the measures of the \(A_j\). The same argument applies if some values are infinite, since all terms are nonnegative. The sets \(S\) and \(N\) partition \(X\). Therefore, for every \(A\subseteq X\), \(\nu(A)=\sum_{x\in A}\nu(\{x\})=\nu_{\mathrm{ac}}(A)+\nu_{\mathrm{s}}(A)\).
If \(\mu(A)=0\), every \(x\in A\) has \(\mu(\{x\})=0\), since \(\{x\}\subseteq A\) and measures are monotone. Hence \(A\subseteq N\), so \(\nu_{\mathrm{ac}}(A)=0\). This proves \(\nu_{\mathrm{ac}}\ll\mu\). Also \(\nu_{\mathrm{s}}(X\setminus N)=0\) and \(\mu(N)=0\), proving \(\nu_{\mathrm{s}}\perp\mu\).
For \(x\in S\), both singleton masses in the ratio defining \(f(x)\) are finite, and the denominator is positive, so \(f(x)\) is finite and nonnegative. On \(N\), \(f\) is zero. Every function on \((X,\mathcal{P}(X))\) is measurable. To verify the integral identity, first note that for any nonnegative measurable function on a countable space, integration is the sum of its singleton contributions: \(\int_A f\,d\mu=\sum_{x\in A}f(x)\mu(\{x\})\). This follows by applying the Monotone Convergence Theorem to the increasing finite partial sums of the nonnegative point-mass contributions (and is also the integral formula for a discrete measure). At \(x\in S\), \(f(x)\mu(\{x\})=\nu(\{x\})\); at \(x\in N\), this product is zero. Consequently, \(\int_A f\,d\mu=\sum_{x\in A\cap S}\nu(\{x\})=\nu_{\mathrm{ac}}(A)\), as required. If \(X\) is empty, all sums are empty and equal zero, and the conclusions still hold. \(\square\)
The density is specified only up to \(\mu\)-almost-everywhere equality, as usual for Radon–Nikodym derivatives. Its chosen value \(0\) on \(N\) is convenient, but the values there do not affect any integral because \(\mu(N)=0\).
Worked Example: A Finite Space with Both Parts
Let \(X=\{a,b,c,d\}\), with singleton masses \(\mu(\{a\})=2,\ \mu(\{b\})=0,\ \mu(\{c\})=3,\ \mu(\{d\})=0\), and \(\nu(\{a\})=6,\ \nu(\{b\})=5,\ \nu(\{c\})=0,\ \nu(\{d\})=4\). Both measures are finite, hence sigma-finite. Find the decomposition and a density.
The reference-supported points are \(S=\{a,c\}\), since their \(\mu\)-masses are \(2>0\) and \(3>0\). The reference-null points are \(N=\{b,d\}\), whose \(\mu\)-masses are both zero. The formula gives \(\nu_{\mathrm{ac}}(\{a\})=6,\ \nu_{\mathrm{ac}}(\{c\})=0\), with zero absolutely continuous mass at \(b,d\). The singular part has masses \(5\) at \(b\) and \(4\) at \(d\), and zero at \(a,c\). The density is \(f(a)=6/2=3,\ f(c)=0/3=0,\ f(b)=0,\ f(d)=0\).
For a check on \(A=\{a,b,c\}\), the original measure is \(\nu(A)=6+5+0=11\). The integral of the density is \(\int_A f\,d\mu=3\cdot2+0\cdot0+0\cdot3=6\), and the singular part is \(\nu_{\mathrm{s}}(A)=5\). Thus \(11=6+5\). On all of \(X\), \(\nu(X)=6+5+0+4=15\), while \(\nu_{\mathrm{ac}}(X)=6\) and \(\nu_{\mathrm{s}}(X)=5+4=9\), giving \(15=6+9\).
Worked Example: A Weighted Reference Measure with a Null Point
Take \(X=\mathbb{N}\cup\{\infty\}\). Define \(\mu(\{n\})=2^{-n}\) for \(n\ge1\), \(\mu(\{\infty\})=0\), and define \(\nu(\{n\})=3\cdot4^{-n}\) for \(n\ge1\), \(\nu(\{\infty\})=7\). These measures are finite: the geometric sums give \(\mu(X)=\sum_{n\ge1}2^{-n}=1\) and \(\nu(X)=\sum_{n\ge1}3\cdot4^{-n}+7=1+7=8\).
Here \(S=\mathbb{N}\) and \(N=\{\infty\}\). Thus \(\nu_{\mathrm{ac}}(\{n\})=3\cdot4^{-n}\) for \(n\ge1\), while \(\nu_{\mathrm{s}}(\{\infty\})=7\). The density at a supported point is \(f(n)=(3\cdot4^{-n})/(2^{-n})=3\cdot2^{-n}\), and \(f(\infty)=0\).
For a set check, let \(A=\{2,4,6,\ldots\}\). Then \(\nu_{\mathrm{ac}}(A)=\sum_{k\ge1}3\cdot4^{-2k}=\sum_{k\ge1}3\cdot16^{-k}=3(1/16)/(1-1/16)=1/5\). The density integral gives \(\int_A f\,d\mu=\sum_{k\ge1}(3\cdot2^{-2k})2^{-2k}=\sum_{k\ge1}3\cdot16^{-k}=1/5\). The geometric-series computation verifies both the density and the exact value independently. The singular mass on \(A\) is zero because \(\infty\notin A\).
Worked Example: Counting Measure Has No Null Points
Let \(X=\mathbb{N}\), let \(\mu\) be counting measure, and set \(\nu(\{n\})=2^{-n}\) for every \(n\ge1\). Counting measure is sigma-finite, since \(\mathbb{N}\) is a countable union of finite sets; \(\nu\) is finite because \(\sum_{n\ge1}2^{-n}=1\). Every singleton has \(\mu\)-mass \(1>0\), so \(S=X\) and \(N\) is empty. The decomposition therefore has \(\nu_{\mathrm{ac}}=\nu\) and \(\nu_{\mathrm{s}}=0\), with \(f(n)=2^{-n}/1=2^{-n}\).
For \(A=\{1,2,3\}\), the direct measure is \(\nu(A)=2^{-1}+2^{-2}+2^{-3}=1/2+1/4+1/8=7/8\). The integral is the same sum, because integration against counting measure sums the function values: \(\int_A f\,d\mu=2^{-1}+2^{-2}+2^{-3}=7/8\). On the whole space, \(\nu(X)=\sum_{n\ge1}2^{-n}=1=\int_X f\,d\mu\), confirming the decomposition.
Why the Null Points Must Be Separated First
A common mistake is to try to define the density by dividing singleton masses at every point. In the weighted example, at \(\infty\) this would give \(\nu(\{\infty\})/\mu(\{\infty\})=7/0\), which is not a finite real number and does not define the density there. The correction is not to treat that quotient as an infinite density: \(\infty\) belongs to \(N\), its mass is assigned to \(\nu_{\mathrm{s}}\), and the density of \(\nu_{\mathrm{ac}}\) is set to zero there. Since \(\mu(\{\infty\})=0\), this chosen value contributes zero to the integral.
This separation also explains the absolute-continuity test directly. A set of zero \(\mu\)-measure can contain no point of \(S\), so the absolutely continuous part assigns it zero mass. Conversely, the singular part is carried entirely by \(N\), a set of zero \(\mu\)-measure. The two parts are determined by the reference measure, not by whether the original measure has atoms: the relevant question is whether each point has positive mass under \(\mu\).
For uniqueness, suppose another pair of positive measures \(\alpha,\beta\) satisfies \(\nu=\alpha+\beta\), \(\alpha\ll\mu\), and \(\beta\perp\mu\). Choose \(N_0\) with \(\mu(N_0)=0\) and \(\beta(X\setminus N_0)=0\). No point of \(S\) can belong to \(N_0\), because then \(0<\mu(\{x\})\le\mu(N_0)=0\), a contradiction. Thus \(\beta(\{x\})=0\) for \(x\in S\). For \(x\in N\), \(\mu(\{x\})=0\), so absolute continuity gives \(\alpha(\{x\})=0\). Since \(\nu=\alpha+\beta\), these facts force \(\alpha(\{x\})=\nu(\{x\})\) on \(S\) and \(\beta(\{x\})=\nu(\{x\})\) on \(N\). Every subset of \(X\) is a countable union of singletons, so countable additivity forces \(\alpha=\nu_{\mathrm{ac}}\) and \(\beta=\nu_{\mathrm{s}}\). This proves uniqueness on the countable space. \(\square\)
Check Your Understanding
In each question, assume the underlying countable set has the sigma-algebra of all subsets.
- Let \(\mu(\{a\})=4,\ \mu(\{b\})=0\), and \(\nu(\{a\})=10,\ \nu(\{b\})=3\). Find \(S,N,\nu_{\mathrm{ac}},\nu_{\mathrm{s}}\), and a density of \(\nu_{\mathrm{ac}}\) with respect to \(\mu\).
- Why does sigma-finiteness imply that every singleton has finite measure? Give the argument for an arbitrary point of the countable space.
- For counting measure on \(\mathbb{N}\) and \(\nu(\{n\})=3^{-n}\), determine the singular part and calculate \(\nu(\{1,2\})\) from the density.
- Explain why a point with \(\mu(\{x\})=0\) and \(\nu(\{x\})>0\) cannot contribute a positive amount to \(\nu_{\mathrm{ac}}\).
- Prove directly from singleton masses that any two decompositions into an absolutely continuous part and a part singular with respect to \(\mu\) must agree.