When Do Set Probabilities Determine a Law?
A probability law assigns a probability to every measurable event, but in applications we often know probabilities only for a structured family of events. For a real-valued random variable, the probabilities of intervals \((-\infty,x]\) are its cumulative distribution function. The key question is why these probabilities determine the law on every Borel set. The answer is a uniqueness principle: agreement on a generating pi-system forces agreement on the sigma-algebra it generates.
All measures in the main theorem are probability measures on the same measurable space \((X,\mathcal{M})\). They are finite, with total mass \(1\); no completeness or sigma-finiteness assumption is needed. A pi-system is a family of sets closed under finite intersections. A lambda-system is a family containing \(X\), closed under complements relative to \(X\), and closed under countable disjoint unions. These closure properties are different: a lambda-system need not be a sigma-algebra.
A sigma-algebra is a lambda-system, but the converse does not follow from the definitions alone: a lambda-system is only required to contain countable unions when the sets are disjoint. The pi-lambda theorem explains how the two types of closure work together. We prove it here because it is the mechanism behind the uniqueness result.
The Pi-Lambda Theorem and Uniqueness
Proof. The smallest lambda-system \(\mathcal{D}\) containing \(\mathcal{P}\) exists: it is the intersection of all lambda-systems containing \(\mathcal{P}\), a nonempty collection because \(\mathcal{M}\) is one. First note that any lambda-system is closed under differences of nested members. If \(C,A\in\mathcal{D}\) and \(C\subseteq A\), then \(A^c\) and \(C\) are disjoint members, so \(A^c\cup C\in\mathcal{D}\). Taking its complement gives \(A\setminus C\in\mathcal{D}\).
Fix \(A\in\mathcal{P}\) and define \(\mathcal{D}_A=\{B\in\mathcal{D}:A\cap B\in\mathcal{D}\}\). This is a lambda-system. It contains \(X\) because \(A\cap X=A\in\mathcal{D}\). If \(B\in\mathcal{D}_A\), then \(A\cap B\subseteq A\), so \(A\cap B^c=A\setminus(A\cap B)\in\mathcal{D}\) by the nested-difference property; also \(B^c\in\mathcal{D}\). Thus \(B^c\in\mathcal{D}_A\). Finally, if \(B_n\in\mathcal{D}_A\) are pairwise disjoint, then \(A\cap B_n\) are pairwise disjoint members of \(\mathcal{D}\), and their union is \(A\cap\bigcup_n B_n\). Hence \(\bigcup_n B_n\in\mathcal{D}_A\). For every \(B\in\mathcal{P}\), the pi-system property gives \(A\cap B\in\mathcal{P}\subseteq\mathcal{D}\); therefore \(\mathcal{D}_A\) contains \(\mathcal{P}\). By minimality, \(\mathcal{D}\subseteq\mathcal{D}_A\). We have proved \(A\cap B\in\mathcal{D}\) for all \(A\in\mathcal{P}\) and \(B\in\mathcal{D}\).
Now fix \(B\in\mathcal{D}\), and define \(\mathcal{H}_B=\{A\in\mathcal{D}:A\cap B\in\mathcal{D}\}\). It is a lambda-system: \(X\cap B=B\in\mathcal{D}\); complements follow from \(A^c\cap B=B\setminus(A\cap B)\) and nested differences; and disjoint unions follow by intersecting each set with \(B\) and using closure under disjoint unions. The result of the preceding paragraph shows that \(\mathcal{P}\subseteq\mathcal{H}_B\). Minimality gives \(\mathcal{D}\subseteq\mathcal{H}_B\). Consequently, \(\mathcal{D}\) is closed under finite intersections.
Closure under complements and finite intersections gives closure under finite unions. For any sequence \(A_n\in\mathcal{D}\), set \(C_1=A_1\) and, for \(n\ge2\), set \(C_n=A_n\setminus\bigcup_{j=1}^{n-1}A_j\). Finite unions and nested differences show \(C_n\in\mathcal{D}\); the \(C_n\) are pairwise disjoint and \(\bigcup_n C_n=\bigcup_n A_n\). Thus \(\mathcal{D}\) is a sigma-algebra. It contains \(\mathcal{P}\), so \(\sigma(\mathcal{P})\subseteq\mathcal{D}\). Conversely, \(\sigma(\mathcal{P})\) is a lambda-system containing \(\mathcal{P}\), so minimality gives \(\mathcal{D}\subseteq\sigma(\mathcal{P})\). The two families are equal. \(\square\)
Proof. Let \(\mathcal{D}=\{E\in\sigma(\mathcal{P}):\mu(E)=\nu(E)\}\). Since both measures are probabilities, \(\mu(X)=\nu(X)=1\), so \(X\in\mathcal{D}\). If \(E\in\mathcal{D}\), finite total mass and additivity give \(\mu(E^c)=1-\mu(E)=1-\nu(E)=\nu(E)\), so \(E^c\in\mathcal{D}\). If \(E_n\in\mathcal{D}\) are pairwise disjoint, countable additivity gives \(\mu(\bigcup_n E_n)=\sum_n\mu(E_n)=\sum_n\nu(E_n)=\nu(\bigcup_n E_n)\); thus their union lies in \(\mathcal{D}\). Therefore \(\mathcal{D}\) is a lambda-system. It contains \(\mathcal{P}\) by the assumed agreement. The Pi-Lambda Theorem implies \(\sigma(\mathcal{P})\subseteq\mathcal{D}\), which is the required equality. \(\square\)
The probability assumption matters in this proof: it supplies equality on \(X\) and lets complements be handled by subtracting from total mass \(1\). If a pi-system used in an application does not contain \(X\), adjoin \(X\) to it. The enlarged family is still a pi-system and generates the same sigma-algebra; agreement on \(X\) is automatic for probability measures.
Worked Example: Initial-Interval Probabilities Determine a Law on [0,1]
Let \(P\) be a probability measure on \(([0,1],\mathcal{B}_{[0,1]})\) satisfying \(P([0,t])=t\) for every \(t\in[0,1]\). Show that \(P\) is Lebesgue measure restricted to \([0,1]\), denoted \(Q\).
Take \(\mathcal{P}=\{\varnothing\}\cup\{[0,t]:0\le t\le1\}\). It contains \([0,1]\), and \([0,s]\cap[0,t]=[0,\min(s,t)]\), so it is a pi-system. It generates \(\mathcal{B}_{[0,1]}\): it contains the initial intervals, their complements give \((t,1]\), and differences give \((s,t]\) for \(0\le s<t\le1\). These intervals generate the relative open intervals, and relative open intervals generate the Borel sigma-algebra. The restricted Lebesgue measure \(Q\) is a probability measure because \(m([0,1])=1\), and \(Q([0,t])=m([0,t])=t\). Also \(P(\varnothing)=Q(\varnothing)=0\). Thus \(P\) and \(Q\) agree on \(\mathcal{P}\), and the uniqueness theorem gives \(P(E)=Q(E)\) for every Borel set \(E\subseteq[0,1]\).
Check: at \(t=0\), both assign probability \(0\) to \([0,0]=\{0\}\); at \(t=1\), both assign probability \(1\) to \([0,1]\). For \(0\le s<t\le1\), additivity on the disjoint union \([0,t]=[0,s]\cup(s,t]\) gives \(P((s,t])=t-s\), matching \(Q((s,t])=m((s,t])=t-s\).
Worked Example: Recovering an Exponential Law from Its CDF
Suppose \(P\) is a probability measure on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\) with \(P((-\infty,x])=0\) for \(x<0\) and \(P((-\infty,x])=1-e^{-x}\) for \(x\ge0\). Show that it equals the probability measure \(Q\) with density \(e^{-x}\chi_{[0,\infty)}(x)\) relative to Lebesgue measure.
The measure \(Q\) has total mass \(\int_0^\infty e^{-x}\,dx=1\). Let \(\mathcal{P}=\{\varnothing,\mathbb{R}\}\cup\{(-\infty,x]:x\in\mathbb{R}\}\). It is a pi-system because the intersection of two such half-lines is the one with the smaller endpoint. It generates \(\mathcal{B}_{\mathbb{R}}\): for example, \((-\infty,b)=\bigcup_{n\ge1}(-\infty,b-1/n]\), and open intervals are intersections of sets of the forms \((-\infty,b)\) and \((a,\infty)\). For \(x<0\), \(Q((-\infty,x])=0\); for \(x\ge0\), direct integration gives \(Q((-\infty,x])=\int_0^x e^{-u}\,du=1-e^{-x}\). So \(P\) and \(Q\) agree on \(\mathcal{P}\), and the theorem gives \(P=Q\) on all Borel sets.
For a numerical interval check, \(P((1,2])=P((-\infty,2])-P((-\infty,1])=(1-e^{-2})-(1-e^{-1})=e^{-1}-e^{-2}\). The density gives \(Q((1,2])=\int_1^2e^{-u}\,du=[-e^{-u}]_1^2=e^{-1}-e^{-2}\), as required.
Worked Example: Lower-Left Rectangles Determine a Joint Law
Let \(Q\) be the probability measure on \((\mathbb{R}^2,\mathcal{B}_{\mathbb{R}^2})\) defined by \(Q(E)=m_2(E\cap[0,1]^2)\), where \(m_2\) is Lebesgue measure on the plane. Suppose another probability measure \(P\) satisfies, for all real \(a,b\),
The right side is the area of \([0,1]^2\cap((-\infty,a]\times(-\infty,b])\): the horizontal side length is \(\max(0,\min(a,1))\), and the vertical side length is \(\max(0,\min(b,1))\). Thus it equals the corresponding \(Q\)-probability. The family of lower-left rectangles, together with \(\varnothing\) and \(\mathbb{R}^2\), is a pi-system: intersections use the smaller endpoint in each coordinate. It generates \(\mathcal{B}_{\mathbb{R}^2}\), since coordinate half-planes can be obtained by countable unions of these rectangles, and open rectangles form a basis for the topology. The uniqueness theorem therefore gives \(P=Q\).
For \(a=\tfrac12\) and \(b=\tfrac34\), the formula gives \(P((-\infty,\tfrac12]\times(-\infty,\tfrac34])=\tfrac12\cdot\tfrac34=\tfrac38\). Independently, the intersection with the unit square is a rectangle with area \(\tfrac12\cdot\tfrac34=\tfrac38\), so the value agrees with \(Q\). If \(a=-1\) and \(b=\tfrac34\), the first side length is \(\max(0,\min(-1,1))=0\), and both measures assign probability \(0\).
A Common Mistake: Agreement Is Not Enough
The family must generate the sigma-algebra on which equality is claimed. For a concrete counterexample, take \(X=\{0,1\}\), \(\mathcal{M}=\mathcal{P}(X)\), and let \(\mu\) put mass \(1\) at \(0\), while \(\nu\) puts mass \(1\) at \(1\). On the family \(\{\varnothing,X\}\), both measures assign probabilities \(0\) and \(1\), respectively. But \(\sigma(\{\varnothing,X\})=\{\varnothing,X\}\), not \(\mathcal{M}\), and \(\mu(\{0\})=1\ne0=\nu(\{0\})\). Thus agreement on this family does not imply agreement on all measurable sets.
A related error is to apply the theorem to a collection that is not a pi-system or to forget that the total masses must agree. The proof relies on closure under intersections when establishing the Pi-Lambda Theorem, and on equal total mass when taking complements. For probability measures the latter condition is built in; for finite measures of different total mass it is not. The practical test is to verify both the generating property and the required closure properties before using uniqueness.
Check Your Understanding
For each question, identify the sigma-algebra on which the conclusion is meant to hold and check the hypotheses of the uniqueness theorem.
- Why is the family \(\{(-\infty,x]:x\in\mathbb{R}\}\cup\{\varnothing,\mathbb{R}\}\) a pi-system?
- Suppose two probability measures on \([0,1]\) agree on every set \([0,t]\), \(0\le t\le1\). What generated sigma-algebra does the uniqueness theorem give?
- In the proof of uniqueness, which property of probability measures ensures that agreement passes from \(E\) to \(E^c\)?
- Give two distinct probability measures on \(\{0,1\}\) that agree on \(\{\varnothing,\{0,1\}\}\), and identify why the uniqueness theorem does not imply they agree on every subset.
- For the exponential law in the worked example, compute the probability of \((2,3]\) using both the CDF values and the density integral.