What a Cumulative Difference Measures
A cumulative function records how much measure lies to the left of each real number, including that number. The difference between two cumulative values therefore records precisely the measure added between their cutoffs. The inclusion convention matters: with the definition used here, the interval recovered is \((a,b]\), not an interval with both endpoints included or both excluded.
Throughout, let \((\mathbb{R},\mathcal{B}_{\mathbb{R}},\mu)\) be a finite Borel measure space: \(\mu\) is a nonnegative countably additive measure on the Borel sigma-algebra and \(\mu(\mathbb{R})<\infty\). It is consequently sigma-finite, but we do not assume it is complete. Define the real-valued cumulative function \(F\) at every \(x\in\mathbb{R}\) by
Because \(\mu\) is finite, \(F(x)\) is finite for every \(x\). It is nondecreasing: if \(x\le y\), then \((-\infty,x]\subseteq(-\infty,y]\), so measure monotonicity gives \(F(x)\le F(y)\). If \(\mu\) is a probability measure, this is the usual cumulative distribution function. The formula below applies more generally to any finite Borel measure, whether or not its total mass is one, and does not require a density.
Proof. Suppose first that \(a<b\). The sets \((-\infty,a]\) and \((a,b]\) are disjoint, and their union is \((-\infty,b]\). By finite additivity, which follows from countable additivity, \(\mu((-\infty,b])=\mu((-\infty,a])+\mu((a,b])\). Substituting the definition of \(F\) gives \(F(b)=F(a)+\mu((a,b])\). Both cumulative values are finite, so subtraction is valid and yields the formula. If \(a=b\), then \((a,b]=(a,a]=\varnothing\), so its measure is zero, and \(F(b)-F(a)=F(a)-F(a)=0\). \(\square\)
This identity is an exact recovery rule: once a measure and its cumulative function are given, no separate integration is needed to find the mass of \((a,b]\). It also confirms that every such increment is nonnegative. Indeed, \(F(b)-F(a)=\mu((a,b])\ge0\), in agreement with the fact that \(F\) is nondecreasing. The converse direction—starting with a candidate function and constructing a measure from it—is a separate question.
Partitions and Telescoping Increments
The formula remains useful when an interval is divided into smaller pieces. Each piece must use the same endpoint convention: \((x_{j-1},x_j]\). This makes adjacent pieces disjoint and ensures that their union is the original half-open interval.
Proof. For each \(j\), the interval \((x_{j-1},x_j]\) contains points strictly greater than \(x_{j-1}\) and at most \(x_j\). These intervals are pairwise disjoint, and their union over \(j=1,\ldots,n\) is \((a,b]\). Finite additivity therefore gives the first equality. Applying the interval increment formula to each interval gives the second equality. The sum of differences is \((F(x_1)-F(x_0))+\cdots+(F(x_n)-F(x_{n-1}))\); every intermediate value \(F(x_j)\), for \(1\le j<n\), occurs once with a plus sign and once with a minus sign. The sum is \(F(x_n)-F(x_0)=F(b)-F(a)\), as required. \(\square\)
The partition formula is a consistency check as well as a calculation method. If computed interval masses fail to add up across an adjacent partition, then either a cumulative value or an endpoint convention has been used incorrectly.
Worked Example: A Discrete Measure
Let \(\mu=2\delta_{-1}+3\delta_{1/2}+\delta_3\), where \(\delta_c\) is the Dirac measure assigning mass one to a Borel set containing \(c\) and mass zero to a set not containing \(c\). Thus the three points \(-1,\frac12,3\) have respective masses \(2,3,1\), and \(\mu(\mathbb{R})=2+3+1=6\).
To calculate \(F(x)=\mu((-\infty,x])\), include a point mass exactly when its location is at most \(x\). Consequently, \(F(x)=0\) for \(x<-1\); \(F(x)=2\) for \(-1\le x<\frac12\); \(F(x)=5\) for \(\frac12\le x<3\); and \(F(x)=6\) for \(x\ge3\). The endpoint inequalities follow from the cumulative definition’s use of \(\le x\).
Take \(a=-1\) and \(b=\frac12\). The interval \((-1,\frac12]\) excludes \(-1\) and includes \(\frac12\), so its only mass is \(3\). The cumulative increment gives the same answer:
Check: direct membership gives \(0\) from the point \(-1\), \(3\) from \(\frac12\), and \(0\) from \(3\), for a total of \(0+3+0=3\). For the larger interval \((-1,3]\), the formula gives \(F(3)-F(-1)=6-2=4\); direct membership includes \(\frac12\) and \(3\), whose masses sum to \(3+1=4\).
Worked Example: A Uniform Measure on a Bounded Interval
Define a probability measure on \(\mathbb{R}\) by \(\mu(E)=\frac14\int_E\chi_{[-2,2]}(x)\,dm(x)\) for Borel sets \(E\), where \(m\) is Lebesgue measure. This measure has constant density \(\frac14\) on \([-2,2]\) and zero outside it. Its cumulative function is
For \(-2\le x\le2\), the cumulative mass is the density \(\frac14\) times the length \(x-(-2)=x+2\). At \(x=-2\), that length is zero. At \(x=2\), it is \(4\), giving total mass \(1\); beyond \(2\), the cumulative mass stays \(1\).
Calculate the mass of \((-1,\frac32]\). Both endpoints lie in \([-2,2]\), so \(F(\frac32)=\frac{3/2+2}{4}=\frac{7/2}{4}=\frac78\), while \(F(-1)=\frac{-1+2}{4}=\frac14\). Hence
Check: the interval has length \(\frac32-(-1)=\frac52\), so its density-based mass is \(\frac14\cdot\frac52=\frac58\). The cumulative calculation agrees with this independent calculation.
Worked Example: An Exponential Probability Measure
Let \(\mu\) be the probability measure on \(\mathbb{R}\) with density \(e^{-x}\chi_{[0,\infty)}(x)\) with respect to Lebesgue measure. For \(x<0\), its cumulative mass is zero. For \(x\ge0\), integrating the density gives \(F(x)=\int_0^x e^{-t}\,dt=1-e^{-x}\). The total mass is \(\int_0^\infty e^{-t}\,dt=1\), so \(\mu\) is finite and \(F\) takes values in \([0,1]\).
For \(a=\ln2\) and \(b=\ln4\), we have \(0<a<b\), because \(1<2<4\) and the logarithm is increasing. The cumulative values are \(F(\ln4)=1-e^{-\ln4}=1-\frac14=\frac34\) and \(F(\ln2)=1-e^{-\ln2}=1-\frac12=\frac12\). Therefore
Check: direct integration gives \(\int_{\ln2}^{\ln4}e^{-x}\,dx=e^{-\ln2}-e^{-\ln4}=\frac12-\frac14=\frac14\). Including or excluding either individual endpoint does not alter this integral because a singleton has Lebesgue measure zero, but the increment formula itself still specifies \((\ln2,\ln4]\).
Endpoint Conventions and a Common Error
The formula is not a license to ignore the brackets. It follows from the particular decomposition \((-\infty,b]=(-\infty,a]\mathbin{\dot\cup}(a,b]\), where the dot indicates disjoint union. A different interval convention requires separate endpoint accounting; one should not silently substitute another interval for \((a,b]\).
Common mistake: using the increment for a closed interval. Let \(\mu=2\delta_0+3\delta_{1/2}\). Then \(F(0)=2\) and \(F(\frac12)=5\), so the increment is \(F(\frac12)-F(0)=5-2=3\). This is \(\mu((0,\frac12])\): it excludes \(0\) and includes \(\frac12\). It is not \(\mu([0,\frac12])\), which includes both points and has mass \(2+3=5\). The correction is to match the interval to the formula: use the increment for \((0,\frac12]\), or separately account for the mass at \(0\) when the desired interval is closed. Thus the interval notation and the cumulative-function convention must be checked before subtracting.
Another immediate check is the sign. For \(a\le b\), \(F(b)-F(a)\) is nonnegative because it equals a measure. Reversing the subtraction generally gives a nonpositive number and cannot be the mass of \((a,b]\), except when that mass is zero. The formula applies to \(a\le b\); it is not an instruction to treat \((a,b]\) as a nonempty interval when \(a>b\).
In applications, cumulative functions are valuable because one function encodes all half-open interval masses at once. The formula works for discrete measures, absolutely continuous measures, and measures with neither a density nor a finite list of point masses. For a finite push-forward measure \(\nu\) on \(\mathbb{R}\), the same result applies with \(F(x)=\nu((-\infty,x])\); once that cumulative function is known, interval masses of the push-forward follow by subtraction.
Check Your Understanding
In each question, keep the interval endpoints and the definition \(F(x)=\mu((-\infty,x])\) explicit.
- For a finite Borel measure \(\mu\), prove directly that \(\mu((a,b])=F(b)-F(a)\) when \(a=b\).
- If \(F(1)=\frac25\) and \(F(4)=\frac9{10}\) for a finite measure, what is \(\mu((1,4])\)? What property guarantees that your answer cannot be negative?
- Let \(\mu=4\delta_{-2}+\delta_0+2\delta_5\). Compute \(\mu((-2,5])\) by finding the two needed cumulative values, and verify by checking which points belong to the interval.
- For \(a=x_0<x_1<x_2=b\), write the two increment terms whose sum gives \(\mu((a,b])\), and show that the intermediate cumulative value cancels.
- For the measure with density \(\frac14\chi_{[-2,2]}\), compute \(\mu((-2,0])\) from the cumulative function and check it by integrating the density over the interval.