Distribution functions are monotone right-continuous encodings of measures on the real line. Lebesgue-Stieltjes measures make this correspondence exact.
Distribution functions
A distribution function records mass accumulated from the left.
Definition. A distribution function is an increasing right-continuous function \(F:\mathbb R\to\mathbb R\) with finite limits at \(\pm\infty\). For probability laws, these limits are \(0\) and \(1\).
Proposition. If \(\mu\) is a finite Borel measure on \(\mathbb R\), then \(F(x)=\mu((-\infty,x])\) is increasing and right-continuous.
Proof. Monotonicity is immediate from inclusion. If \(x_n\downarrow x\), then \((-\infty,x_n]\downarrow(-\infty,x]\), and continuity from above for finite measures gives \(F(x_n)\downarrow F(x)\).
Measure generated by a distribution function
Right-continuity is the convention that fixes endpoint masses.
Theorem. For every distribution function \(F\), there is a unique finite Borel measure \(\mu_F\) such that \(\mu_F((a,b])=F(b)-F(a)\).
Proof. On bounded half-open intervals define \[ \mu_0((a,b])=F(b)-F(a), \] and set \(\mu_0((-\infty,b])=F(b)-F(-\infty)\). Extend by finite additivity to the algebra of finite disjoint unions. If \(I_n\downarrow\varnothing\) are algebra sets, monotonicity of \(F\), right-continuity at finite endpoints, and its limits at infinity imply \(\mu_0(I_n)\downarrow0\). This continuity at the empty set is equivalent to countable additivity for a finite premeasure. Carath?odory extension gives \(\mu_F\) on \(\mathcal B(\mathbb R)\). Half-lines form a generating \(\pi\)-system and the measure is finite, so uniqueness follows from the \(\pi\)-\(\lambda\) theorem.
Proposition (bijection). The assignments \[ \mu\longmapsto F_\mu(x)=\mu((-\infty,x]), \qquad F\longmapsto\mu_F \] are inverse bijections between finite Borel measures on \(\mathbb R\) and increasing right-continuous functions with finite limits, after fixing the left limit \(F(-\infty)=0\).
Proof. Starting from \(\mu\), interval subtraction gives \[ F_\mu(b)-F_\mu(a)=\mu((a,b]), \] so uniqueness of the Stieltjes extension yields \(\mu_{F_\mu}=\mu\). Starting from \(F\), construction gives \(\mu_F((-\infty,x])=F(x)-F(-\infty)\), recovering the normalized \(F\).
Atoms and jumps
Point masses are exactly jumps of the distribution function.
Proposition. \(\mu_F(\{x\})=F(x)-F(x-)\), where \(F(x-)=\lim_{y\uparrow x}F(y)\).
Proof. The singleton \(\{x\}\) is the decreasing intersection of intervals \((x-1/n,x]\). Continuity from above gives \(\mu_F(\{x\})=\lim_n(F(x)-F(x-1/n))=F(x)-F(x-)\).
Integration against \(F\)
Stieltjes notation is integration with respect to \(\mu_F\).
Definition. For measurable \(g\), write \(\int g\,\mathrm{d}F\) for \(\int g\,\mathrm{d}\mu_F\), whenever the latter integral is defined.
Remark. In probability, if \(X\) has distribution function \(F\), then \[ \mathbb E g(X)=\int_{\mathbb R}g\,\mathrm{d}F. \] If \(F\) has density \(p\), this is \(\int gp\,\mathrm{d}x\); if \(F\) is purely discrete with masses \(p_j\) at \(x_j\), it is \(\sum_jg(x_j)p_j\). Mixed and singular continuous laws require no change in notation.
Weak convergence via distribution functions
Continuity points are the right places to test convergence.
Theorem. For probability measures on \(\mathbb R\), \(\mu_n\Rightarrow\mu\) if and only if \(F_n(x)\to F(x)\) at every continuity point of \(F\).
Proof. If \(\mu_n\Rightarrow\mu\), the portmanteau inequalities applied to the closed set \((-\infty,x]\) and open set \((-\infty,x)\) give \[ F(x-)\le\liminf_nF_n(x)\le\limsup_nF_n(x)\le F(x). \] At a continuity point \(F(x-)=F(x)\), so \(F_n(x)\to F(x)\).
Conversely, first choose continuity points \(a<b\) with \(\mu((a,b])>1-\varepsilon\). Partition \([a,b]\) by continuity points \(a=x_0<\cdots<x_m=b\) so finely that a bounded continuous \(g\) oscillates by at most \(\varepsilon\) on each interval. Distribution-function convergence gives \[ \mu_n((x_{j-1},x_j])\to\mu((x_{j-1},x_j]) \] for every \(j\), while the two tails have asymptotically small mass. Upper and lower step approximations to \(g\) then show \(\int g\,\mathrm{d}\mu_n\to\int g\,\mathrm{d}\mu\).