Monotonicity, Right Continuity, and Limits of Cumulative Functions

What the Cumulative Function Records

A cumulative function records how much measure lies to the left of a moving threshold. Its defining sets are nested as the threshold moves, which gives monotonicity. Right continuity and the limits at the two ends of the real line follow from continuity properties of measures; the finiteness assumption is essential for the right-continuity argument.

Let \(\nu\) be a finite measure on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), where finite means \(\nu(\mathbb{R})<\infty\). Write \(M=\nu(\mathbb{R})\), allowing \(M=0\). Define \(F:\mathbb{R}\to[0,M]\) by the measure of the closed half-line ending at \(t\). If \(\nu\) comes from a measurable map \(T:(X,\mathcal{M})\to(\mathbb{R},\mathcal{B}_{\mathbb{R}})\) and a finite measure \(\mu\) on \((X,\mathcal{M})\), its push-forward is \(\nu(B)=\mu(T^{-1}(B))\) for \(B\in\mathcal{B}_{\mathbb{R}}\). In that case the same cumulative function can be computed from the preimage of the half-line.

The equality involving \(T\) is pointwise in \(t\); it follows from the definition of push-forward and the Borel measurability of \((-\infty,t]\). No probability normalization is assumed: \(M\) need not be \(1\). In the results below, statements about a general finite Borel measure \(\nu\) also apply to such a push-forward.

The Measure Continuity Used in the Proof

We will use continuity from below and continuity from above. These are consequences of countable additivity. In the second statement, the finiteness of the first set's measure is required.

Proof. For the increasing sequence, set \(D_1=A_1\) and \(D_n=A_n\setminus A_{n-1}\) for \(n\ge2\). The \(D_n\) are disjoint, \(\bigcup_nD_n=\bigcup_nA_n\), and \(A_N=\bigcup_{n=1}^N D_n\). Countable additivity gives \(\lambda(\bigcup_n A_n)=\sum_{n=1}^{\infty}\lambda(D_n)\), while finite additivity gives \(\lambda(A_N)=\sum_{n=1}^N\lambda(D_n)\). The partial sums increase to the series, proving continuity from below. For the decreasing sequence, the sets \(B_n=A_1\setminus A_n\) increase to \(B=A_1\setminus\bigcap_n A_n\). Since \(\lambda(A_1)<\infty\), subtraction is valid and \(\lambda(A_n)=\lambda(A_1)-\lambda(B_n)\). Continuity from below gives \(\lambda(B_n)\to\lambda(B)\); subtracting from \(\lambda(A_1)\) yields \(\lambda(A_n)\to\lambda(\bigcap_nA_n)\). \(\square\)

Monotonicity, Right Continuity, and Tail Limits

Proof. If \(s\le t\), then \((-\infty,s]\subseteq(-\infty,t]\). Monotonicity of the measure therefore gives \(F(s)\le F(t)\).

Fix \(t\in\mathbb{R}\), and let \(t_n=t+1/n\). The sets \((-\infty,t_n]\) decrease, and their intersection is \((-\infty,t]\): a real number \(x\) belongs to every one of them exactly when \(x\le t\). Since \(\nu((-\infty,t_1])\le M<\infty\), continuity from above gives \(F(t_n)\to F(t)\). To verify right continuity for arbitrary thresholds, let \(\varepsilon>0\). Choose \(N\) such that \(F(t+1/N)<F(t)+\varepsilon\), using this sequence limit. Whenever \(t<s<t+1/N\), monotonicity gives \(F(t)\le F(s)\le F(t+1/N)<F(t)+\varepsilon\). Thus \(F(s)\to F(t)\) as \(s\) approaches \(t\) from the right.

For the limit at minus infinity, the sets \((-\infty,-n]\) decrease to the empty set as \(n\to\infty\). Their first member has finite measure, so continuity from above gives \(F(-n)\to\nu(\varnothing)=0\). Given \(\varepsilon>0\), choose \(N\) with \(F(-N)<\varepsilon\). For every \(t\le -N\), monotonicity gives \(0\le F(t)\le F(-N)<\varepsilon\), proving the limit as \(t\to-\infty\).

For the limit at plus infinity, the sets \((-\infty,n]\) increase to \(\mathbb{R}\). Continuity from below gives \(F(n)\to\nu(\mathbb{R})=M\). Given \(\varepsilon>0\), choose \(N\) with \(M-F(N)<\varepsilon\). If \(t\ge N\), then \(F(N)\le F(t)\le M\), so \(0\le M-F(t)\le M-F(N)<\varepsilon\). This proves the limit as \(t\to+\infty\). If \(M=0\), monotonicity and \(0\le F(t)\le M\) give \(F(t)=0\) for every \(t\), and the conclusions hold as well. \(\square\)

Computing the Cumulative Function

Why Finiteness Matters

The proof of right continuity used continuity from above for decreasing sets whose first set has finite measure. That condition is not automatic for an infinite measure. For a concrete example, define the Borel measure \(\nu=\sum_{n=1}^{\infty}\delta_{1/n}\), which assigns unit mass to each point \(1/n\). It is not finite: \(\nu(\mathbb{R})=\infty\). At \(t=0\), no point \(1/n\) is at most \(0\), so \(F(0)=0\). For every \(t>0\), infinitely many points \(1/n\) satisfy \(1/n\le t\), so \(F(t)=\infty\). In particular, \(F(1/k)=\infty\) for each positive integer \(k\), although \(1/k\downarrow0\); thus the right limit at \(0\) does not equal \(F(0)\).

A common mistake is to apply continuity from above to a decreasing sequence without checking that its first set has finite measure. For \(A_k=(-\infty,1/k]\) in this example, \(A_k\downarrow(-\infty,0]\), but every \(A_k\) has infinite measure, while the intersection has measure \(0\). The finite-measure hypothesis is exactly what licenses the continuity-from-above step in the theorem.

For a finite measure, the three properties have distinct sources: nested half-lines give monotonicity, decreasing half-lines and finite-measure continuity from above give right continuity, and continuity from below or above gives the two tail limits. The total mass need not be one, but it is always the limit at plus infinity.