Singular Continuous Laws and the Cantor Example

Why a Continuous Law Need Not Have a Density

A familiar continuous distribution assigns probabilities by integrating a density. But “continuous” can also mean that the distribution function has no jumps, or equivalently that no single point has positive probability. Those properties alone do not guarantee that a density exists. The Cantor distribution is a concrete example: it has no point masses, yet all its probability is concentrated on a set of length zero.

The key is to keep two questions separate. Does the law assign positive probability to any individual point? And can probabilities be obtained by integrating a function over intervals? For the Cantor distribution, the answer to the first question is no, while the answer to the second is also no.

Constructing the Cantor Distribution

Let \(B_1,B_2,\ldots\) be independent random variables with \(P(B_k=0)=P(B_k=1)=1/2\) for every \(k\). Define

The series converges for every sequence of zeroes and ones, because its terms are nonnegative and

This random variable chooses, with equal probability, between a left and a right third at each stage. To make that description precise, after fixing the first \(n\) variables the possible values of \(X\) lie in an interval

because the remaining tail is between zero and \(\sum_{k=n+1}^{\infty}2/3^k=1/3^n\). Each particular choice of the first \(n\) variables has probability \(2^{-n}\).

At stage \(n\), the set \(C_n\) consists of \(2^n\) closed intervals, each of length \(3^{-n}\). The first \(n\) choices of the \(B_k\)'s determine which of these intervals contains \(X\). Therefore \(X\in C_n\) for every \(n\), and hence \(P(X\in C)=1\).

No Point Masses, but Probability One on a Null Set

The construction provides direct proofs of the two features that make this law unusual. First, the Cantor set has length zero. Second, no single point can carry positive probability.

Proof. The total length of the \(2^n\) intervals in \(C_n\) is \(2^n3^{-n}=(2/3)^n\). Since \(C\subseteq C_n\) for every \(n\), its Lebesgue measure satisfies \[ \lambda(C)\leq (2/3)^n \] for every \(n\). Letting \(n\) increase gives \(\lambda(C)=0\).

Now fix any real number \(x\). At stage \(n\), the intervals making up \(C_n\) are disjoint and separated, so at most one of them contains \(x\). If none contains \(x\), then \(P(X=x)=0\), since \(X\in C_n\) with probability one. If one interval contains \(x\), the event \(X=x\) is contained in the event that the first \(n\) Bernoulli variables choose that interval. The latter event has probability \(2^{-n}\). Thus \(P(X=x)\leq 2^{-n}\) for every \(n\), and letting \(n\) increase yields \(P(X=x)=0\). \(\square\)

The law is therefore concentrated on a set of Lebesgue measure zero while assigning zero probability to each point. These are not contradictory claims: the Cantor set contains uncountably many points. A probability law can spread its total probability across such a set without placing positive probability at any one of its members.

The Distribution Function and Its Self-Similarity

The first Bernoulli variable determines whether \(X\) falls in the left or right third. More precisely, if \(B_1=0\), then \(X=X'/3\); if \(B_1=1\), then \(X=(2+X')/3\), where \(X'=\sum_{j=1}^{\infty}2B_{j+1}/3^j\) has the same distribution as \(X\) and is independent of \(B_1\). This gives the following recursive description of \(F\).

To verify the first recursive piece, take \(0\leq x\leq 1/3\). Conditional on \(B_1=0\), the event \(X\leq x\) is the event \(X'\leq 3x\), of probability \(F(3x)\). Conditional on \(B_1=1\), \(X\geq 2/3\), so it cannot contribute in this range. Multiplying the first conditional probability by \(P(B_1=0)=1/2\) gives \(F(x)=F(3x)/2\).

For \(1/3\leq x\leq 2/3\), all outcomes with \(B_1=0\) are included and outcomes with \(B_1=1\) are excluded, except possibly a boundary value at \(2/3\). That boundary has probability zero by the no-point-mass result. Thus \(F(x)=1/2\) throughout the middle third. For \(2/3\leq x\leq1\), the left branch contributes \(1/2\), and the right branch contributes \(F(3x-2)/2\), giving the final recursive piece. The endpoint formulas agree because \(F(0)=0\) and \(F(1)=1\).

The middle-third formula shows that the distribution function is flat on \([1/3,2/3]\). This is consistent with the construction: the open middle third contains no possible values of \(X\), and neither endpoint has a point mass. The same pattern recurs inside the remaining thirds.

Why the CDF Is Continuous

For any distribution function, the size of its jump at \(x\) is \(P(X=x)\). Since the Cantor distribution has no point masses, its distribution function has no jumps. Distribution functions are right-continuous, and their left-hand limit at \(x\) is \(P(X<x)\); here that limit equals \(F(x)\) because \(P(X=x)=0\). Thus \(F\) is continuous everywhere.

This continuity does not make \(F\) an integral of a density. In fact, outside \(C\), the function is locally constant: every point outside the Cantor set lies in a removed gap or outside \([0,1]\). The complement of \(C\) has full Lebesgue measure, so \(F\) has derivative zero almost everywhere. Yet \(F(1)-F(0)=1\). The familiar recovery of changes in a CDF by integrating its derivative applies when the CDF is absolutely continuous; continuity alone is not enough.

Three Types of Behavior to Keep Distinct

The Cantor distribution illustrates a third behavior alongside discrete and density-based laws. A discrete law assigns positive probability to one or more individual points. A law with a density assigns probabilities by integration and gives probability zero to every Lebesgue-null set. The Cantor law has neither feature: it has no point masses, but it is concentrated on a null set, so it cannot have an ordinary density.

A common mistake is to argue, “The CDF is continuous and has no jumps, so its derivative is a density.” The correction is that differentiability and continuity are not the same as absolute continuity. Even where a derivative exists almost everywhere, it need not recover the total change in the CDF by integration. In this example the derivative is zero almost everywhere, while the distribution still has total probability one.