Why Cumulative Values Determine a Measure
For a finite Borel measure on the real line, the cumulative distribution function records the measure of every half-line \((-\infty,x]\). These values also determine the measure of intervals: for \(a<b\), the mass of \((a,b]\) is the difference between the cumulative values at \(b\) and \(a\). The uniqueness result goes further: agreement on these half-lines forces agreement on every Borel set. Its proof rests on a generating \(\pi\)-system and the \(\pi\)-\(\lambda\) theorem.
Throughout, \(\mu\) and \(\nu\) are finite, nonnegative Borel measures on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\). Finite means \(\mu(\mathbb{R})<\infty\) and \(\nu(\mathbb{R})<\infty\). Their cumulative distribution functions are the real-valued functions \(F_\mu(x)=\mu((-\infty,x])\) and \(F_\nu(x)=\nu((-\infty,x])\), defined for every real \(x\). No assumption that the measures have total mass one is needed.
These are distinct closure conditions. In particular, a \(\lambda\)-system need not be a \(\sigma\)-algebra. The \(\pi\)-\(\lambda\) theorem connects the two types of families.
The \(\pi\)-\(\lambda\) Theorem
Proof. Let \(\mathcal{L}\) be the intersection of all \(\lambda\)-systems on \(X\) that contain \(\mathcal{P}\). Such systems exist because the power set of \(X\) is one, and their intersection is again a \(\lambda\)-system. Thus \(\mathcal{L}\) is the smallest \(\lambda\)-system containing \(\mathcal{P}\), and \(\mathcal{L}\subseteq\mathcal{D}\).
We first note that a \(\lambda\)-system is closed under differences \(B\setminus A\) when \(A\subseteq B\) and \(A,B\) belong to it. Indeed, \(A\) and \(B^c\) are disjoint members, so \(A\cup B^c\) belongs to the system; its complement is \(B\setminus A\).
Fix \(A\in\mathcal{P}\), and define \(\mathcal{L}_A=\{B\in\mathcal{L}:A\cap B\in\mathcal{L}\}\). Since \(A\in\mathcal{L}\), this family contains \(X\). If \(B\in\mathcal{L}_A\), then \(A\cap B^c=A\setminus(A\cap B)\in\mathcal{L}\) by the difference property, so \(B^c\in\mathcal{L}_A\). If the sets \(B_n\in\mathcal{L}_A\) are pairwise disjoint, then the sets \(A\cap B_n\) are pairwise disjoint members of \(\mathcal{L}\), and their union \(A\cap\bigcup_nB_n\) belongs to \(\mathcal{L}\). Hence \(\mathcal{L}_A\) is a \(\lambda\)-system. It contains \(\mathcal{P}\), since \(A\cap B\in\mathcal{P}\) for \(B\in\mathcal{P}\). By minimality, \(\mathcal{L}\subseteq\mathcal{L}_A\). Consequently \(A\cap B\in\mathcal{L}\) whenever \(A\in\mathcal{P}\) and \(B\in\mathcal{L}\).
Now fix \(B\in\mathcal{L}\), and define \(\mathcal{C}_B=\{A\in\mathcal{L}:A\cap B\in\mathcal{L}\}\). The same complement, disjoint-union, and \(X\)-membership arguments show that \(\mathcal{C}_B\) is a \(\lambda\)-system. The preceding paragraph shows it contains \(\mathcal{P}\), so minimality gives \(\mathcal{L}\subseteq\mathcal{C}_B\). Thus \(\mathcal{L}\) is closed under finite intersections. Together with complements, this gives finite unions. For any sequence \(A_n\in\mathcal{L}\), the sets \(A_1\) and \(A_n\setminus\bigcup_{k<n}A_k\), for \(n\ge2\), are pairwise disjoint members of \(\mathcal{L}\); their union is \(\bigcup_nA_n\). Therefore \(\mathcal{L}\) is a \(\sigma\)-algebra. It contains \(\mathcal{P}\), so \(\sigma(\mathcal{P})\subseteq\mathcal{L}\subseteq\mathcal{D}\), as required. \(\square\)
Half-Lines Generate the Borel Sets
Set \(\mathcal{P}=\{(-\infty,x]:x\in\mathbb{R}\}\). It is a \(\pi\)-system because the intersection of the members indexed by \(x\) and \(y\) is \((-\infty,\min(x,y)]\). Each member is closed, hence Borel. Conversely, for any real \(a\), density of the rationals gives \((-\infty,a)=\bigcup_{q\in\mathbb{Q},\,q<a}(-\infty,q]\), so this open ray belongs to \(\sigma(\mathcal{P})\). Also \((a,\infty)=(-\infty,a]^c\). Intersecting these rays shows that every open interval \((a,b)\) belongs to \(\sigma(\mathcal{P})\). Every open subset of \(\mathbb{R}\) is a countable union of open intervals with rational endpoints contained in it: the rational intervals form a countable base for the usual topology. Thus every open set belongs to \(\sigma(\mathcal{P})\). Since the Borel sigma-algebra is generated by the open sets, \(\sigma(\mathcal{P})=\mathcal{B}_{\mathbb{R}}\).
Uniqueness from Cumulative Distribution Functions
Proof. First, the assumed equality gives equality of the total masses. The sets \((-\infty,n]\) increase to \(\mathbb{R}\) as positive integers \(n\) tend to infinity. Continuity from below for measures therefore gives \(\mu(\mathbb{R})=\lim_{n\to\infty}F_\mu(n)=\lim_{n\to\infty}F_\nu(n)=\nu(\mathbb{R})\).
Define \(\mathcal{D}=\{E\in\mathcal{B}_{\mathbb{R}}:\mu(E)=\nu(E)\}\). This family contains \(\mathbb{R}\) by equality of total masses. If \(E\in\mathcal{D}\), finiteness permits subtraction from the total masses: \(\mu(E^c)=\mu(\mathbb{R})-\mu(E)=\nu(\mathbb{R})-\nu(E)=\nu(E^c)\). Thus \(E^c\in\mathcal{D}\). If \(E_n\in\mathcal{D}\) are pairwise disjoint, countable additivity gives \(\mu(\bigcup_nE_n)=\sum_n\mu(E_n)=\sum_n\nu(E_n)=\nu(\bigcup_nE_n)\), so their union belongs to \(\mathcal{D}\). Hence \(\mathcal{D}\) is a \(\lambda\)-system. The hypothesis says it contains \(\mathcal{P}\), and the \(\pi\)-\(\lambda\) theorem gives \(\mathcal{B}_{\mathbb{R}}=\sigma(\mathcal{P})\subseteq\mathcal{D}\). Therefore the measures agree on every Borel set. \(\square\)
Worked Example: Detecting Different Atomic Measures
Let \(\mu=\frac13\delta_{-1}+\frac23\delta_2\) and \(\nu=\frac13\delta_{-1}+\frac23\delta_3\), finite Borel measures on \(\mathbb{R}\). Here \(\delta_c\) assigns mass one to a Borel set containing \(c\), and zero otherwise. The cumulative values are
For example, at \(x=2\), the half-line includes the atom at \(2\) but not the atom at \(3\). Thus \(F_\mu(2)=\frac13+\frac23=1\), whereas \(F_\nu(2)=\frac13\). The measures are unequal, as the uniqueness theorem's contrapositive requires. Directly, \(\mu([2,3))=\frac23\) and \(\nu([2,3))=0\), confirming the difference. The endpoint convention matters: the jump at \(2\) is included in \(F_\mu(2)\).
Worked Example: A Nontrivial Pushforward with Uniform Distribution
Let \(m\) be Lebesgue measure on \(\mathbb{R}\), and define \(T:[0,1]\to[0,1]\) by \(T(t)=2t\) for \(0\le t\le\frac12\), and \(T(t)=2-2t\) for \(\frac12\le t\le1\). This continuous map defines a finite Borel measure \(\nu\) on \(\mathbb{R}\) by \(\nu(B)=m(\{t\in[0,1]:T(t)\in B\})\). Compare it with \(\mu(B)=m(B\cap[0,1])\), also a finite Borel measure.
For \(x<0\), neither measure assigns mass to \((-\infty,x]\). For \(0\le x\le1\), the preimage of \((-\infty,x]\) consists of \([0,x/2]\) and \([1-x/2,1]\). These intervals have lengths \(x/2\) and \(x/2\); if \(x=1\), they share only the point \(1/2\), which has measure zero. Hence \(F_\nu(x)=x\). For \(x>1\), the preimage is all of \([0,1]\), of measure \(1\). The same cases give \(F_\mu(x)=0\) for \(x<0\), \(F_\mu(x)=x\) for \(0\le x\le1\), and \(F_\mu(x)=1\) for \(x>1\). The theorem now proves \(\nu(B)=\mu(B)\) for every Borel set \(B\), not only for half-lines.
Check: at \(x=\frac14\), the two preimage lengths are \(\frac18\) and \(\frac18\), so \(F_\nu(\frac14)=\frac14=F_\mu(\frac14)\). At \(x=1\), the two lengths are \(\frac12\) and \(\frac12\), so each cumulative value is \(1\). As an independent check on a set, the preimage of \((\frac14,\frac34]\) has length \((\frac38-\frac18)+(\frac78-\frac58)=\frac12\), matching \(m((\frac14,\frac34]\cap[0,1])=\frac12\).
Why Finiteness Cannot Be Dropped
Finiteness is used precisely when passing from \(E\) to \(E^c\): equality of total mass and of the mass of \(E\) must allow subtraction. The failure is not merely technical. Let \(\mu=m\) and \(\nu=2m\) on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\). Both are infinite measures. For every real \(x\), \(\mu((-\infty,x])=\infty=\nu((-\infty,x])\), but \(\mu([0,1])=1\) and \(\nu([0,1])=2\). Thus agreement of extended cumulative values does not imply uniqueness for arbitrary infinite measures.
Common mistake. One might define \(\mathcal{D}=\{E:\mu(E)=\nu(E)\}\) for these two infinite measures and claim it is a \(\lambda\)-system because it contains every half-line. That claim fails: the set \(E=(-\infty,0]\cup[2,\infty)\) has \(\mu(E)=\nu(E)=\infty\), but \(E^c=(0,2)\) has \(\mu(E^c)=2\) and \(\nu(E^c)=4\). So \(\mathcal{D}\) is not closed under complements. The incorrect step is effectively treating \(\infty-\infty\) as a defined subtraction; it is not. In the finite-measure theorem, equal finite total masses make the complement calculation valid.
Check Your Understanding
Use the definitions and proof structure above to answer these questions.
- Verify that the intersection of \((-\infty,a]\) and \((-\infty,b]\) is again a member of the half-line \(\pi\)-system.
- Explain why equality of all cumulative values implies equality of the total masses of two finite measures.
- For \(\mu=\delta_0\) and \(\nu=\delta_1\), find a value of \(x\) at which their cumulative distribution functions differ, and compute both values.
- Identify the exact step in the uniqueness proof that can fail for measures with infinite total mass.
- Show that every open interval \((a,b)\) belongs to the sigma-algebra generated by the half-lines \((-\infty,x]\).