Cumulative Distribution Functions of Finite Measures

Cumulative Mass up to a Threshold

A measure on the real line assigns mass to Borel sets, while a cumulative function records how much mass lies at or to the left of a chosen threshold. The threshold convention matters: the set used here is $(-\infty,x]$, so a point mass located exactly at $x$ is included. For a finite measure, this cumulative value is always a finite real number.

Throughout, let $(\mathbb{R},\mathcal{B}_{\mathbb{R}},\mu)$ be a finite Borel measure space. Thus $\mu$ is defined on the Borel sigma-algebra $\mathcal{B}_{\mathbb{R}}$ and $M=\mu(\mathbb{R})<\infty$. In particular, this measure space is sigma-finite, since $\mathbb{R}$ itself has finite measure. We do not assume that the Borel measure space is complete. The function defined below is real-valued and its defining equality holds at every real $x$, not merely almost everywhere.

Definition: The cumulative distribution function, or cumulative function, of $\mu$ is $F:\mathbb{R}\to[0,M]$ given by $$ F(x)=\mu((-\infty,x]),\qquad x\in\mathbb{R}. $$

The set $(-\infty,x]$ is Borel, so the value is defined. Also, $(-\infty,x]\subseteq\mathbb{R}$, and therefore $0\le F(x)\le\mu(\mathbb{R})=M$. A finite measure need not have total mass one: when $M=1$, it is a probability measure, but the definition does not require that normalization. The calculation is always a set-membership calculation: include precisely the mass at locations $a$ satisfying $a\le x$.

Point Masses and Finite Atomic Measures

For $a\in\mathbb{R}$, the Dirac measure $\delta_a$ is the Borel measure defined by $\delta_a(E)=1$ if $a\in E$, and $\delta_a(E)=0$ if $a\notin E$, for every $E\in\mathcal{B}_{\mathbb{R}}$. A weighted point mass $c\delta_a$, where $c\ge0$, has total mass $c$. Its cumulative value changes according to whether the threshold has reached $a$.

Worked Example: A Single Point Mass

Let $\mu=5\delta_2$ on $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$. This is a finite Borel measure with total mass $\mu(\mathbb{R})=5$. For any real $x$, the set $(-\infty,x]$ contains the point $2$ exactly when $2\le x$. Thus $$ F(x)= \begin{cases} 0,&x<2,\\ 5,&x\ge2. \end{cases} $$ In particular, $F(1)=0$, $F(2)=5$, and $F(3)=5$. Check: at the threshold $x=2$, the defining set is $(-\infty,2]$, which contains $2$, so the value is $5$; at $x=1$, it does not contain $2$, so the value is $0$.

The same membership rule gives a formula for any finite atomic measure. Such a measure is a finite sum of weighted Dirac measures. Locations in that sum need not be distinct; if two locations coincide, their weights both contribute when the threshold includes that location.

Proposition: Let $N$ be a nonnegative integer, let $a_1,\ldots,a_N\in\mathbb{R}$, and let $c_1,\ldots,c_N\ge0$. On $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$, define the finite atomic measure $\mu=\sum_{j=1}^{N}c_j\delta_{a_j}$, with the empty sum interpreted as the zero measure when $N=0$. Then, for every $x\in\mathbb{R}$, $$ F(x)=\sum_{\substack{1\le j\le N\\a_j\le x}}c_j. $$

Proof. By the definition of the cumulative function, $F(x)=\mu((-\infty,x])$. By the definition of the finite sum of measures, this equals $\sum_{j=1}^{N}c_j\delta_{a_j}((-\infty,x])$. For each index $j$, $\delta_{a_j}((-\infty,x])=1$ if $a_j\le x$, and it equals $0$ if $a_j>x$. Removing the zero terms gives the displayed sum. If $N=0$, the sum is empty and equals $0$, which agrees with the zero measure of $(-\infty,x]$. $\square$

Worked Example: A Finite Atomic Measure

Let $\mu=2\delta_{-1}+3\delta_1+4\delta_3$. This is a finite Borel measure with total mass $2+3+4=9$. For $x<-1$, none of the locations $-1,1,3$ satisfies $a_j\le x$, so $F(x)=0$. For $-1\le x<1$, only $-1\le x$, so $F(x)=2$. For $1\le x<3$, both $-1\le x$ and $1\le x$, so $F(x)=2+3=5$. For $x\ge3$, all three locations are included, so $F(x)=2+3+4=9$. Therefore $$ F(x)= \begin{cases} 0,&x<-1,\\ 2,&-1\le x<1,\\ 5,&1\le x<3,\\ 9,&x\ge3. \end{cases} $$ For example, $F(0)=2$, since $-1\le0$ but $1\not\le0$, and $F(1)=2+3=5$, since the atom at $1$ is included at equality. Check: at $x=3$, the included weights are $2,3,4$, so $F(3)=9=\mu(\mathbb{R})$; at $x=2$, the included weights are $2,3$, so $F(2)=5$.

Uniform Measure on an Interval

For real numbers $a<b$, uniform probability measure on the closed interval $[a,b]$ is the finite Borel measure on $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$ defined by $$ \mu(E)=\frac{m(E\cap[a,b])}{b-a},\qquad E\in\mathcal{B}_{\mathbb{R}}, $$ where $m$ is Lebesgue measure on $\mathbb{R}$. Its total mass is $m([a,b])/(b-a)=(b-a)/(b-a)=1$. The cumulative value is the normalized length of the part of $[a,b]$ that lies at or to the left of $x$. Although the interval is closed, its individual endpoints have Lebesgue measure zero, so including or excluding a single endpoint does not change this length. The formula below follows directly by considering thresholds to the left of the interval, inside it, and to its right.

Proposition: If $a<b$ and $\mu(E)=m(E\cap[a,b])/(b-a)$ on $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$, then the cumulative function is $$ F(x)= \begin{cases} 0,&x<a,\\ \dfrac{x-a}{b-a},&a\le x<b,\\ 1,&x\ge b. \end{cases} $$

Proof. By definition, $F(x)=m((-\infty,x]\cap[a,b])/(b-a)$. If $x<a$, the intersection is empty, so its Lebesgue measure and $F(x)$ are both zero. If $a\le x<b$, the intersection is $[a,x]$. Its Lebesgue measure is $x-a$, including when $x=a$, when that length is zero. Hence $F(x)=(x-a)/(b-a)$ in this range. If $x\ge b$, the intersection is all of $[a,b]$, whose Lebesgue measure is $b-a$; dividing by $b-a>0$ gives $F(x)=1$. These cases cover every real $x$, with the boundary $x=b$ handled by the last case. $\square$

Worked Example: Uniform Measure on an Interval

Take $[a,b]=[-2,4]$, so the length is $4-(-2)=6$. The uniform probability measure is $\mu(E)=m(E\cap[-2,4])/6$, and its cumulative function is $$ F(x)= \begin{cases} 0,&x<-2,\\ \dfrac{x+2}{6},&-2\le x<4,\\ 1,&x\ge4. \end{cases} $$ At $x=-3$, the intersection with $[-2,4]$ is empty, giving $F(-3)=0$. At $x=-2$, the intersection has length $-2-(-2)=0$, giving $F(-2)=0/6=0$. At $x=1$, the intersection is $[-2,1]$, of length $1-(-2)=3$, giving $F(1)=3/6=1/2$. At $x=4$, the intersection has length $4-(-2)=6$, giving $F(4)=6/6=1$. At $x=5$, it remains the full interval, so $F(5)=6/6=1$. Check: each value is the intersection length divided by $6$; in particular the threshold $x=4$ includes the right endpoint and yields the full mass $1$.

Threshold Conventions and a Common Pitfall

The notation $(-\infty,x]$ fixes the boundary convention throughout: equality $a_j=x$ counts an atom at $a_j$. Replacing this set with $(-\infty,x)$ would define a different cumulative quantity at thresholds that coincide with atoms. For measures with no point mass at the threshold, that change may leave the value unchanged, but it cannot be made in general.

Worked Example: Correcting a Strict-Inequality Error

Let $\mu=2\delta_{-1}+3\delta_1+4\delta_3$, and evaluate at $x=1$. A common incorrect calculation uses only locations satisfying $a_j<1$, obtaining $2$ from the atom at $-1$ and omitting the weight $3$ at $1$. That calculation gives the mass of $(-\infty,1)$, not the cumulative function defined here. The correct condition is $a_j\le1$, so the locations $-1$ and $1$ both contribute: $$ F(1)=2+3=5. $$ Check: the defining set is $(-\infty,1]$; it contains both $-1$ and $1$, but not $3$, so the included weights sum to $2+3=5$.

A second source of mistakes is treating every finite measure as a probability measure. For the atomic measure in the preceding example, the total mass is $9$, and the cumulative value at $x=3$ is $9$, not $1$. The uniform measure on an interval was normalized by its length, so its total mass is $1$; a different finite measure need not use that normalization. To compute $F(x)$, first use the measure as given, then evaluate the mass of exactly $(-\infty,x]$.

These computations provide direct examples of how a measure is summarized by accumulated mass, without requiring a density. For a finite collection of point masses, the calculation is a finite sum of the weights whose locations meet the threshold. For uniform measure on an interval, it is an intersection length divided by the interval length. In each setting, checking equality at the threshold and checking the total mass are useful safeguards against endpoint and normalization errors.

Check Your Understanding

For each question, use the defining set $(-\infty,x]$ and state explicitly whether a location equal to $x$ is included.

For $\mu=4\delta_0$, write $F(x)$ on the two ranges separated by $0$, and compute $F(0)$.
For $\mu=\delta_{-2}+5\delta_2$, compute $F(-2)$, $F(0)$, and $F(2)$.
For uniform probability measure on $[1,5]$, compute $F(0)$, $F(3)$, and $F(5)$ from intersection lengths.
Give the total mass of $\mu=2\delta_1+3\delta_4$, and compute its cumulative value at $x=4$.
Explain why using $a_j<x$ instead of $a_j\le x$ gives the wrong cumulative value at an atom, using a finite atomic measure of your choice.