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The Cantor distribution is the law of a random variablewhere the $B_n$ are independent Bernoulli random variables taking values zero and one with equal probability. Its values lie in the Cantor set, and its cumulative distribution function is the Cantor function. Every individual value has probability zero, but the Cantor set has probability one despite having zero Lebesgue measure. A probability density function on that set would integrate to zero, so it cannot describe this law. The example motivates the distinction between a continuous probability distribution and an absolutely continuous probability distribution.

The Cantor set is obtained from $[0,1]$ by repeatedly deleting the open middle third of every remaining interval. Its first stages retain $[0,1/3]\cup [2/3,1]$, then four intervals of length $1/9$, then eight of length $1/27$. After level $n$, there are $2^n$ retained intervals of length $3^{-n}$. The Cantor set is the intersection of all these stages, rather than the set of endpoints visible in any finite drawing.

Equivalently,Thus its points admit a base-three expansion using only zero and two. The word “admit” handles numbers with two expansions: $1/3=0.1_3=0.02222{\dots }_3$ is retained as an endpoint. Every infinite binary choice determines a point of $C$, and differing choices give differing points, so $C$ is uncountable.

Its total retained interval length is $(2/3{)}^n\to 0$, so $C$ has zero Lebesgue measure. It is nevertheless nonempty, compact, and uncountable, and has Hausdorff dimension $\log 2/\log 3$: the covers and mass bounds are worked out in Hausdorff dimension of the Cantor set. This separates counting, ordinary length, and fine-scale geometric size in one concrete example.

The self-similarity is exact:The Cantor distribution puts half the probability into each copy at every split. It has a continuous cumulative distribution function, the Cantor function, while all its mass lies on a Lebesgue null set. This makes the construction useful in measure theory as well as fractal geometry: it provides a concrete model in which several plausible meanings of “small” disagree.

A real probability distribution with a continuous cumulative distribution function assigns probability zero to each individual value. Here this is the broad meaning of continuous probability distribution. Many elementary examples are more specifically absolutely continuous probability distributions, described by a probability density function with respect to Lebesgue measure. Some texts use "continuous" for that narrower class. The Cantor distribution shows why the distinction matters: its cumulative distribution function is continuous, yet no probability density function with respect to Lebesgue measure represents its whole law. ^[ref].

The cumulative distribution function of a real random variable $X$ isIt records the probability accumulated up to a threshold. It is a nondecreasing function and is right-continuous; its limits are zero at $-\infty$ and one at $+\infty$. These properties follow from ordered threshold events, continuity from above of a measure, and continuity from below of a measure. A cumulative distribution function determines the entire probability distribution on the real line. ^[ref].

For $a<b$,where $F_X(x^{-})={\lim }_{t\uparrow x}{F}_X(t)$ is its left-hand limit. The second formula measures a jump. Thus $\mathbb{P}(a\le X\le b)=F_X(b)-F_X(a^{-})$; changing an endpoint is harmless only when its point probability is zero.

For the finite example under probability distribution, $F_X(x)$ is $0$ below zero, $0.2$ on $[0,1)$, $0.7$ on $[1,3)$, and $1$ on $[3,\infty )$. In particular $F_X(1)=0.7$, but $F_X(1^{-})=0.2$, so the jump gives $\mathbb{P}(X=1)=0.5$. For the density example, $F_Y(y)=y^2$ on $[0,1]$, and its increase from $1/4$ to $3/4$ is $9/16-1/16=1/2$.

When an absolutely continuous probability distribution has probability density function $f$, its cumulative distribution function is $F(x)={\int }_{-\infty }^{x}f(t)dt$ and $F^{\prime }=f$ almost everywhere. A continuous cumulative distribution function by itself does not guarantee such a density: the Cantor distribution is an important exception.

The middle-third Cantor set has Hausdorff dimensionIts Lebesgue measure is zero, yet $0<{\mathcal{H}}^d(C)<\infty$. These assertions measure different aspects of size: interval length is too coarse to distinguish this set from a null set, while the critical fractional measuring rule captures its remaining structure.

At level $n$, the construction covers $C$ by $2^n$ intervals of diameter $3^{-n}$. Their total $s$-cost isFor $s>d$ it tends to zero, proving the upper dimension bound. At $s=d$ it equals one, giving ${\mathcal{H}}^d(C)\le 1$. For $s<d$ this particular cover's cost grows, but that alone would not prove a lower bound on Hausdorff measure.

For the matching lower bound, use the Cantor distribution, which assigns mass $2^{-n}$ to each level-$n$ Cantor cylinder. An interval of length $r$ with $3^{-(n+1)}<r\le 3^{-n}$ meets at most two level-$n$ construction intervals, since their gaps are at least $3^{-n}$. Its mass is therefore at mostEvery covering set of diameter $r$ in the line lies in an interval of length $r$, so any small cover must have total $d$-cost at least $1/4$. This is the mass distribution principle in its interval form: $1/4\le {\mathcal{H}}^d(C)\le 1$, sufficient to prove the claimed dimension. The constants are convenient bounds, rather than an assertion of the exact critical Hausdorff measure. In the diagram, the right panel is a finite-stage approximation to the Cantor function, showing how probability is shared while the middle third receives none.

The reusable pattern is to obtain an upper bound by constructing covers and a lower bound by distributing mass. The same strategy underlies Cartesian powers of the Cantor set and the dimension formula for a finite similarity iterated function system satisfying the open set condition.

The mass distribution principle converts a diffuse measure into a lower bound on Hausdorff dimension. Let $s>0$ and let a Borel probability measure $\mu$ give a Borel set $E$ full mass. If, for some $C,r_0>0$, every ball centered in $E$ satisfiesthen ${\mathcal{H}}^s(E)\ge 1/C>0$ in the diameter-cost convention, and consequently ${\dim }_{H}E\ge s$.

The idea is to keep unit mass from being crowded into cheap covering pieces. For a piece $U$ of positive diameter $r$ meeting $E$, choose $x\in U\cap E$. A ball of radius $r+\epsilon$ contains $U$; letting $\epsilon \downarrow 0$ gives ${\mu }^{\ast }(U)\le C{r}^s$, where ${\mu }^{\ast }$ is the associated outer measure. Zero-diameter pieces have zero mass by the same estimate. For any sufficiently small cover,Thus every cover costs at least $1/C$. This is the lower-bound counterpart to the explicit covers in scale-restricted Hausdorff content. The Cantor distribution supplies the needed mass in Hausdorff dimension of the Cantor set. ^[ref].

A probability density function represents probabilities as areas under a nonnegative measurable function. For a real random variable with an absolutely continuous probability distribution, it satisfiesfor every Borel set $A$. Its height is probability per unit of the outcome variable, not the probability of that exact outcome. Thus a probability density function may exceed one, while the probability of a singleton is zero. NIST describes the interval-probability interpretation through integration. ^[ref].

For the example $f_Y(y)=2y$ on $[0,1]$, $f_Y(0.8)=1.6$, but $\mathbb{P}(Y=0.8)=0$ and $\mathbb{P}(0.7<Y<0.9)=0.9^2-0.7^2=0.32$. The cumulative distribution function is $F_Y(y)=y^2$ within that interval, connecting density height to the local accumulation rate.

Changing the scale changes the density height. If $Z=100Y$, then $f_Z(z)=f_Y(z/100)/100=z/5000$ on $[0,100]$. At the corresponding value $z=80$, its height is $0.016$, while $\mathbb{P}(70<Z<90)=0.32$ is unchanged. The factor $1/100$ compensates for the wider integration variable. This is a change of variables, not a change in the underlying probabilities.

A probability density function is determined only almost everywhere: changing its values on a Lebesgue measure zero set changes no probabilities. The Cantor distribution has no such representation, and a mixed probability distribution can have both a point mass and a density component, so differentiating its cumulative distribution function does not recover all its mass.