Decomposing Measures with Atomic and Continuous Parts

Three Different Kinds of Measure

A measure can assign mass in several ways at once. A density spreads mass according to Lebesgue measure; a Dirac measure places all its mass at one point; and Cantor measure is concentrated on a Lebesgue-null set without placing positive mass at any individual point. Separating these contributions prevents a common confusion: a measure singular to Lebesgue measure need not consist of point masses.

Throughout, measures in the examples are finite positive measures on $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$, where $m$ denotes Lebesgue measure restricted to the Borel sets. A measure $\alpha$ is absolutely continuous with respect to $m$, written $\alpha\ll m$, if $m(E)=0$ implies $\alpha(E)=0$ for every Borel set $E$. It is singular with respect to $m$, written $\beta\perp m$, if there is a Borel set $S$ with $m(S)=0$ and $\beta(S^c)=0$. A point mass of size $a\geq0$ at $x$ is $a\delta_x$, where $\delta_x(E)=1$ if $x\in E$ and $0$ otherwise.

Definition: A finite measure is point-atomic if it is a sum of point masses. It is atomless if every singleton has measure zero. A singular continuous measure is both singular with respect to Lebesgue measure and atomless. Here “continuous” means atomless; it does not mean absolutely continuous with respect to Lebesgue measure.

In a density term $\int_E f\,dm$, the density $f$ is nonnegative and integrable, and its values on an $m$-null set do not affect the measure. The Radon–Nikodym theorem identifies such a density for a measure absolutely continuous with respect to $m$; its uniqueness is up to equality $m$-almost everywhere. The Lebesgue decomposition theorem separates a finite positive measure into an absolutely continuous and a singular part. The results below make the atomic content of those parts explicit for the measures under consideration.

Point Masses in a Finite Measure

Proposition: Let $\nu$ be a finite positive measure on $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$. The set $D=\{x\in\mathbb{R}:\nu(\{x\})>0\}$ is at most countable. The measure $$ \nu_{\mathrm{at}}(E)=\sum_{x\in D\cap E}\nu(\{x\}) $$ is a measure with $\nu_{\mathrm{at}}(E)\leq\nu(E)$. The difference $\nu_{\mathrm{c}}=\nu-\nu_{\mathrm{at}}$ is a finite positive atomless measure.

Proof. For each positive integer $k$, let $D_k=\{x:\nu(\{x\})\geq 1/k\}$. If $D_k$ contained $N$ distinct points, finite additivity on those disjoint singletons would give $\nu(\mathbb{R})\geq N/k$. Thus $D_k$ is finite. Every $x\in D$ belongs to some $D_k$, because $\nu(\{x\})>0$ implies $1/k\leq\nu(\{x\})$ for some $k$. Hence $D=\bigcup_{k\geq1}D_k$ is countable.

For each Borel $E$, the series defining $\nu_{\mathrm{at}}(E)$ is a sum of nonnegative terms. Countable additivity follows by partitioning the countable index set $D$ across any disjoint sequence of Borel sets and summing the same nonnegative terms in the resulting groups. For any finite subset $F\subseteq D\cap E$, disjointness of its singletons gives $\sum_{x\in F}\nu(\{x\})\leq\nu(E)$. Taking the supremum over such finite $F$ gives $\nu_{\mathrm{at}}(E)\leq\nu(E)$. Therefore $\nu_{\mathrm{c}}(E)=\nu(E)-\nu_{\mathrm{at}}(E)$ is nonnegative. It is countably additive because it is the difference of two finite measures and, for disjoint $E_j$, both series $\sum_j\nu(E_j)$ and $\sum_j\nu_{\mathrm{at}}(E_j)$ converge to finite values. Finally, for every $x\in D$, $\nu_{\mathrm{at}}(\{x\})=\nu(\{x\})$; for $x\notin D$, both singleton measures are zero. Thus $\nu_{\mathrm{c}}(\{x\})=0$ for every $x$, proving atomlessness. $\square$

This proposition extracts all point masses, but it does not say that the remaining measure is absolutely continuous. The residual can be singular continuous. To identify that possibility explicitly, we use the standard Cantor probability measure $\kappa$ on the Borel sets of $\mathbb{R}$. It is supported on the middle-thirds Cantor set $C$, gives each of the $2^n$ closed intervals at stage $n$ mass $2^{-n}$, and has total mass one. The set $C$ has Lebesgue measure zero: it is contained in the stage-$n$ union of intervals, whose total length is $(2/3)^n$, tending to zero. Also $\kappa$ has no point masses. A point outside $C$ has zero mass; a point in $C$ lies in at most two stage-$n$ intervals, so its mass is at most $2\cdot2^{-n}$ for every $n$, which tends to zero. Thus $\kappa$ is singular continuous.

Recognizing the Components of a Mixed Measure

Proposition: Suppose $f\geq0$ is integrable, $a_j>0$ with $\sum_j a_j<\infty$, and $x_j$ are distinct points. Suppose also that $\kappa$ is a finite atomless measure supported on a Borel null set $C$, and $c\geq0$. Then $$ \nu(E)=\int_E f\,dm+\sum_j a_j\delta_{x_j}(E)+c\kappa(E) $$ has absolutely continuous component $\nu_{\mathrm{ac}}(E)=\int_E f\,dm$ and singular component $\nu_{\mathrm{s}}(E)=\sum_j a_j\delta_{x_j}(E)+c\kappa(E)$. The point-atomic component consists exactly of the displayed point masses with their displayed weights; the remaining $c\kappa$ term is atomless.

Proof. If $m(E)=0$, then $\int_E f\,dm=0$, so $\nu_{\mathrm{ac}}\ll m$. The set $S=C\cup\{x_j:j\geq1\}$ is Borel and null, because $C$ is null and a countable set has Lebesgue measure zero. The measure $\nu_{\mathrm{s}}$ is concentrated on $S$, so $\nu_{\mathrm{s}}\perp m$. At a singleton $\{x_j\}$, the density term is zero since $m(\{x_j\})=0$, and the Cantor-type term is zero since $\kappa$ is atomless. Thus $\nu(\{x_j\})=a_j$. At any other singleton all three terms vanish. This proves the assertion about point masses. The absolutely continuous and singular components are unique: if $\nu=\alpha+\beta=\alpha'+\beta'$, with $\alpha,\alpha'\ll m$ and $\beta,\beta'\perp m$, choose null sets supporting $\beta$ and $\beta'$, and take their union $S$. On $S^c$, both singular measures vanish; on $S$, both absolutely continuous measures vanish. Evaluating the two decompositions on $E\cap S$ and $E\cap S^c$ for each Borel $E$ gives $\alpha(E)=\alpha'(E)$ and $\beta(E)=\beta'(E)$. $\square$

Worked Example: A Density Plus Two Point Masses

On $(\mathbb{R},\mathcal{B}_{\mathbb{R}},m)$, define $$ \nu(E)=\int_E 2\chi_{[0,1]}\,dm+3\delta_0(E)+\delta_2(E). $$ The density is $2\chi_{[0,1]}$, whose integral is $2\cdot m([0,1])=2\cdot1=2$. The atomic component is $3\delta_0+\delta_2$, of total mass $3+1=4$. Therefore $\nu(\mathbb{R})=2+4=6$. The singular part is supported on the null set $\{0,2\}$, while the density part is absolutely continuous.

For the set $[0,1]$, the density contributes $2\cdot1=2$. The point $0$ belongs to $[0,1]$ and contributes $3$; the point $2$ does not belong and contributes $0$. Hence $\nu([0,1])=2+3=5$. As a second check, $\nu(\{0\})=0+3+0=3$, since a singleton has Lebesgue measure zero and $2\notin\{0\}$; likewise $\nu(\{2\})=0+0+1=1$. These singleton values recover the two atomic weights exactly.

Worked Example: Adding Cantor Measure

Let $\kappa$ be the Cantor probability measure and set $$ \nu(E)=\int_E 2x\chi_{[0,1]}(x)\,dm+4\delta_{1/2}(E)+3\kappa(E). $$ The density has total integral $\int_0^1 2x\,dx=[x^2]_0^1=1-0=1$. The point mass has total mass $4$, and the Cantor term has total mass $3\kappa(\mathbb{R})=3\cdot1=3$. Thus $\nu(\mathbb{R})=1+4+3=8$. Its absolutely continuous component is the density term. Its singular component is $4\delta_{1/2}+3\kappa$, supported on $C\cup\{1/2\}$, a null set.

The Cantor set is contained in $[0,1]$, where $m(C)=0$, and $1/2\notin C$ because it lies in the middle-third gap. Consequently the density and point-mass terms contribute zero to $\nu(C)$, while the Cantor term contributes $3\kappa(C)=3$. Hence $\nu(C)=3$. At the point $1/3\in C$, the density contributes zero, the Dirac mass at $1/2$ contributes zero, and $\kappa(\{1/3\})=0$; therefore $\nu(\{1/3\})=0$. In contrast, $\nu(\{1/2\})=4$, since the density and atomless Cantor measure both give the singleton zero mass. This checks that the Cantor contribution is singular but does not add point atoms.

Worked Example: A Countable Atomic Part and Two Densities

Consider $$ \nu(E)=\int_E\big(\chi_{[0,1]}+2\chi_{[2,3]}\big)\,dm +\sum_{n=1}^{\infty}\frac{3}{2^n}\delta_{n+3}(E)+2\kappa(E). $$ The two density supports each have length one, so the density integral is $1\cdot1+2\cdot1=3$. The atomic masses sum to $3\sum_{n=1}^{\infty}2^{-n}=3\cdot1=3$. The Cantor term has mass $2\cdot1=2$. Thus the total mass is $3+3+2=8$. The absolutely continuous component is the sum of the two density terms. The singular component is the countable atomic sum together with $2\kappa$, supported on $C\cup\{4,5,6,\ldots\}$, a null set.

Since $C\subset[0,1]$, the density integrates to zero over $C$, and none of the points $n+3$ lies in $C$. It follows that $\nu(C)=2\kappa(C)=2$. At $4=1+3$, the atomic series contributes $3/2^1=3/2$; all other terms give zero on the singleton. Thus $\nu(\{4\})=3/2$. These checks distinguish the singular continuous mass on $C$ from the individual atomic weights.

A Common Misreading

A frequent error is to treat the mass of a Cantor-stage interval as a point mass. For the measure in the preceding example, the leftmost stage-one Cantor interval has $\kappa$-mass $1/2$, so the term $2\kappa$ gives that interval mass $2\cdot(1/2)=1$. It does not follow that $2\kappa(\{1/3\})=1$: the point $1/3$ has measure zero under $\kappa$, as shown by atomlessness. The correct calculation is $\nu(\{1/3\})=0$, while $\nu(C)=2$. Mass on a set and mass at each point of that set are different statements.

When decomposing a sum, first identify the density term and compute its integral; next collect all Dirac terms, combining repeated locations if necessary; finally identify any remaining singular atomless measure by a null set that supports it. A singular remainder is not automatically atomic, and an atomless remainder is not automatically absolutely continuous.

Check Your Understanding

For each question, specify which part is absolutely continuous, which is singular, and which point masses are present.

If $\nu(E)=\int_E 3\chi_{[0,2]}\,dm+2\delta_1(E)$, what are $\nu(\mathbb{R})$ and $\nu([0,2])$?
Why does an atomless measure supported on a Borel null set define a singular continuous measure?
For $\nu=5\delta_0+\kappa$, compute $\nu(\{0\})$ and $\nu(C)$. Explain why these values differ.
Prove that the point-atomic part of a finite measure is supported on a countable set.
Can a singular measure have no point masses? Give an example and state a null set supporting it.

Decomposing Measures with Atomic and Continuous Parts

What You'll Learn