Three Ways a Probability Law Can Place Mass
Probability laws on the real line can look very different when examined at individual points and on sets of Lebesgue measure zero. A discrete law places positive mass on one or more points. A law with a density assigns zero mass to every Lebesgue-null Borel set. A singular law concentrates all its mass on a Borel set of Lebesgue measure zero. These descriptions distinguish different properties: in particular, a law can be singular without having any point masses.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \(X:\Omega\to\mathbb{R}\) be a real-valued random variable measurable for the Borel sigma-algebra \(\mathcal{B}_{\mathbb{R}}\). Its law \(\nu\) is the Borel probability measure defined by \(\nu(B)=\mathbb{P}(X\in B)\) for \(B\in\mathcal{B}_{\mathbb{R}}\). Thus \(\nu(\mathbb{R})=1\). Throughout, \(\nu\) is defined on Borel sets, while \(m\) denotes Lebesgue measure on \(\mathbb{R}\), with Lebesgue sigma-algebra \(\mathcal{L}\). When we compare \(\nu\) with \(m\), we will use Borel sets so that \(\nu\) is always evaluated within its domain.
The definition of \(\nu\ll m\) deliberately asks only about Borel null sets. A Borel law \(\nu\) need not be defined on every Lebesgue-measurable set. Every Lebesgue-null set is contained in a Borel null set, so this Borel-set criterion is the appropriate one for a law specified on \(\mathcal{B}_{\mathbb{R}}\).
A law with a density is specified by a nonnegative Borel-measurable function \(f:\mathbb{R}\to[0,\infty]\) such that \(\int_{\mathbb{R}}f\,dm=1\) and \(\nu(B)=\int_B f\,dm\) for every Borel set \(B\). Such a law is absolutely continuous: if \(m(B)=0\), then \(\int_B f\,dm=0\). Conversely, the Radon–Nikodym theorem says that every Borel probability law \(\nu\ll m\) has such a density, up to equality \(m\)-almost everywhere. This is the standard Radon–Nikodym theorem for the sigma-finite measure \(m\) and finite measure \(\nu\).
What the Categories Do—and Do Not—Say
The categories need not be mutually exclusive. A purely atomic law is singular, because a countable set has Lebesgue measure zero. An absolutely continuous law has no point atoms, because every singleton has Lebesgue measure zero. But singularity does not imply the presence of atoms: the Cantor distribution below is singular continuous. Nor does every law have to fit wholly into just one of the three descriptions. For example, a law can be a mixture of an atomic law, a law with a density, and a singular continuous law.
Proof. Suppose first that \(\nu\ll m\). For any \(x\in\mathbb{R}\), the singleton \(\{x\}\) is Borel and \(m(\{x\})=0\). The definition of absolute continuity therefore gives \(\nu(\{x\})=0\). Now suppose \(\nu\perp m\), and let \(N\) be a Borel set with \(m(N)=0\) and \(\nu(N)=1\). If also \(\nu\ll m\), then \(\nu(N)=0\), contradicting \(\nu(N)=1\). Hence the two properties cannot both hold for a probability law. \(\square\)
Proof. By pure atomicity, there is a countable Borel set \(S\) with \(\nu(S)=1\). Every singleton has Lebesgue measure zero, and countable subadditivity gives \(m(S)=0\). Thus \(S\) is a Borel set of Lebesgue measure zero carrying all the probability mass. This is exactly \(\nu\perp m\). \(\square\)
Worked Examples: Discrete and Uniform Laws
Worked Example: A Three-Point Law
Let \(\Omega=\{a,b,c\}\), let \(\mathcal{F}\) be its power set, and set \(\mathbb{P}(\{a\})=1/2\), \(\mathbb{P}(\{b\})=1/3\), and \(\mathbb{P}(\{c\})=1/6\). Define \(X(a)=0\), \(X(b)=1\), and \(X(c)=2\). The weights sum to \(1/2+1/3+1/6=3/6+2/6+1/6=1\), so this is a probability measure.
For a Borel set \(B\subseteq\mathbb{R}\), the law is \(\nu(B)=\tfrac12\chi_B(0)+\tfrac13\chi_B(1)+\tfrac16\chi_B(2)\). In particular, its point masses are \(\nu(\{0\})=1/2\), \(\nu(\{1\})=1/3\), and \(\nu(\{2\})=1/6\); every other singleton has mass zero. The set \(S=\{0,1,2\}\) is countable and \(\nu(S)=1\), so the law is purely atomic and singular.
Check: for \(B=\{0,2\}\), the formula gives \(\nu(B)=1/2+1/6=2/3\). The same value follows directly from \(\mathbb{P}(X\in B)=\mathbb{P}(\{a,c\})=1/2+1/6=2/3\). Also \(S\) has Lebesgue measure zero because it is a finite union of singletons.
Worked Example: The Uniform Law on an Interval
Let \(X\) have the uniform law on \([1,4]\). Its density is \(f(x)=\tfrac13\chi_{[1,4]}(x)\), since \(\int_{\mathbb{R}} f\,dm=\tfrac13 m([1,4])=\tfrac13\cdot3=1\). Thus for every Borel set \(B\), \(\nu(B)=\int_B f\,dm=\tfrac13m(B\cap[1,4])\). If \(B\) is Borel and \(m(B)=0\), then \(m(B\cap[1,4])=0\), so \(\nu(B)=0\); the law is absolutely continuous.
For the interval \([2,3]\), which lies inside \([1,4]\), the probability is \(\nu([2,3])=\tfrac13m([2,3])=\tfrac13\cdot1=\tfrac13\). At a point \(x\in[1,4]\), \(\nu(\{x\})=\tfrac13m(\{x\})=0\), and outside \([1,4]\) the point mass is also zero. Therefore this law has no point atoms.
Check: \(\nu([1,4])=\tfrac13m([1,4])=\tfrac13\cdot3=1\), as required for a probability law. The interval’s endpoint inclusion does not alter the value, since the endpoints are null sets.
Worked Example: The Cantor Distribution
Worked Example: Singular Without Point Atoms
Use \(([0,1],\mathcal{L},m)\) as a probability space. For \(0\le u<1\) and \(n\ge1\), define the binary digit \(b_n(u)=\lfloor2^nu\rfloor-2\lfloor2^{n-1}u\rfloor\); at \(u=1\), set \(b_n(1)=1\). Each \(b_n\) is measurable and takes values in \(\{0,1\}\). Define the measurable real-valued function \(T(u)=\sum_{n=1}^{\infty}2b_n(u)/3^n\). Measurability follows because the partial sums are measurable and converge pointwise. Let \(\nu\) be the law of \(T\), so \(\nu(B)=m(\{u:T(u)\in B\})\) for every Borel set \(B\).
Let \(C_0=[0,1]\), and form \(C_n\) from \(C_{n-1}\) by removing the open middle third of each interval at stage \(n-1\). The set \(C_n\) is a union of \(2^n\) closed intervals, each of length \(3^{-n}\); hence \(m(C_n)=2^n3^{-n}=(2/3)^n\). The Cantor set is \(C=\bigcap_{n=0}^{\infty}C_n\). Since \(C_n\) decreases, \(m(C_0)=1<\infty\), and \(m(C_n)\to0\), continuity from above for measures gives \(m(C)=0\).
Every sum \(T(u)\) has ternary digits only \(0\) and \(2\). After its first \(n\) digits, the remaining tail lies between \(0\) and \(\sum_{k=n+1}^{\infty}2/3^k=3^{-n}\), so the value lies in the retained interval of \(C_n\) determined by those first \(n\) digits. This holds for every \(n\), including the all-ones sequence at \(u=1\), and therefore \(T(u)\in C\). Thus \(\nu(C)=m([0,1])=1\), while \(m(C)=0\); the law is singular.
To check that it has no point atoms, first note that distinct sequences of ternary digits from \(\{0,2\}\) give distinct sums. If two such sequences first differ at index \(k\), the difference contributed there has magnitude \(2/3^k\), while the largest possible opposing tail has magnitude \(\sum_{j>k}2/3^j=1/3^k\). The tail cannot cancel the first difference. Also, the binary digit sequence \((b_n(u))\) determines \(u\), since \(u=\sum_{n=1}^{\infty}b_n(u)/2^n\), including \(u=1\) by the assigned all-ones sequence. Consequently, for any \(x\in\mathbb{R}\), at most one \(u\) satisfies \(T(u)=x\). A singleton has Lebesgue measure zero, so \(\nu(\{x\})=m(\{u:T(u)=x\})=0\). This proves that the Cantor law is singular continuous.
Check: the law assigns probability \(1\) to \(C\), but \(m(C)=0\). If it had a density \(f\), then \(\nu(C)=\int_C f\,dm=0\), contradicting \(\nu(C)=1\). Thus it cannot be absolutely continuous or have a density.
A Common Misclassification
A frequent error is to reason that a law with no point masses must have a density. The Cantor example disproves this: it has \(\nu(\{x\})=0\) for every \(x\), yet it assigns probability \(1\) to a Borel set of Lebesgue measure zero. Having no point atoms is weaker than absolute continuity. To test absolute continuity, one must check all Borel null sets, not just singletons.
A related error is to conclude that a law gives probability zero to a set merely because the set has Lebesgue measure zero. That conclusion is valid for \(\nu\ll m\), not for every probability law. For the Cantor law the set \(C\) is Borel and \(m(C)=0\), but \(\nu(C)=1\). For the discrete law in the first example, the rational numbers form a Borel null set and contain all three support points, so that law also assigns the rationals probability \(1\).
Keep the distinctions precise: absolute continuity describes how a law treats every Borel Lebesgue-null set; singularity means all mass is concentrated on one such set; and point atoms record mass at individual points. A countably supported law is both atomic and singular, whereas the Cantor distribution is singular but has no point atoms.
Check Your Understanding
For each question, identify whether the claim concerns point masses, absolute continuity, or singularity.
- Show that a Borel probability law with a density has zero mass at every singleton.
- A law is concentrated on \(\{1/n:n\ge1\}\). Explain why it is singular with respect to Lebesgue measure.
- Can a singular probability law have zero mass at every point? Give the example from this tutorial and state the two properties that verify your answer.
- For the uniform law on \([1,4]\), compute the probability of \([2,3]\) from its density.
- Why does \(m(B)=0\) alone not imply \(\nu(B)=0\) for an arbitrary Borel probability law? Give a Borel set \(B\) witnessing failure.