Lebesgue-Stieltjes Measures and Distribution Functions

Collect every finite endpoint appearing among the $I_i$ and $J_j$, and adjoin $-\mathrm{\infty }$ or $+\mathrm{\infty }$ when a left or right ray occurs. This gives an increasing list $p_0<\cdots <p_r$ such that each $I_i$ and each $J_j$ is a disjoint union of consecutive h-cells $C_{\mathrm{\ell }}:=(p_{\mathrm{\ell }-1},p_{\mathrm{\ell }}]$, with the first or last cell possibly a ray.

In the unbounded cases this is exactly the truncation/supremum convention built into The interval set function attached to a nondecreasing right-continuous function. Thus each decomposition gives the same total, namely the sum of the cell values over those $C_{\mathrm{\ell }}$ contained in $E$. So ${\mu }_{0,F}(E)$ is well defined. [step 1.1, L1, algebra]

Refine them to a common endpoint grid as in step 1.1, and sum over the grid cells. The cells belonging to $E\cup G$ are exactly the disjoint union of the cells belonging to $E$ and the cells belonging to $G$, so the corresponding cell sums add:

${\mu }_{0,F}(E\cup G)={\mu }_{0,F}(E)+{\mu }_{0,F}(G)\mathrm{.}$

Together with step 2.1 this proves the claim. [step 2.1, given, algebra] ∎

Indeed, if $E={\bigcup }_{r=1}^m{I}_r$ is a finite disjoint union of h-intervals, then each $E_n\cap I_r$ is again a finite disjoint union of h-intervals, the families $(E_n\cap I_r{)}_n$ are pairwise disjoint, and $I_r={\bigcup }_n(E_n\cap I_r)$. Applying the single-interval case to each $I_r$ and summing finitely gives the general case.

For every $N\in \mathbb{N}$,

${\sum }_{n=0}^N{\mu }_{0,F}(I_n)={\mu }_{0,F}\text{\hspace{0.17em}⁣}\left({\bigcup }_{n=0}^N{I}_n\right)\le {\mu }_{0,F}(E),$

because $E$ is the disjoint union of ${\bigcup }_{n=0}^N{I}_n$ and the remainder $E\setminus {\bigcup }_{n=0}^N{I}_n$, whose ${\mu }_{0,F}$-value is nonnegative. Hence ${\sum }_n{\mu }_{0,F}(I_n)\le {\mu }_{0,F}(E)$. [step 1.1, L1, given, algebra]

Let $\epsilon >0$. Right continuity at $a$ gives $\delta >0$ with $F(a+\delta )-F(a)<\epsilon \mathrm{/}2$. For each $n$, if $I_n$ meets $[a+\delta ,b]$ then its right endpoint is finite; write that endpoint as $v_n$, let $u_n$ be its left endpoint, and choose $w_n>v_n$ with $F(w_n)-F(v_n)<\epsilon 2^{-n-2}$. Then the open intervals $(u_n,w_n)$ cover $[a+\delta ,b]$: every $x$ in that compact interval belongs to ${\bigcup }_n{I}_n=(a,b]$, hence lies in some $I_n$ that must have finite right endpoint and therefore satisfies $x\in (u_n,w_n)$. [given, L2, choose]

Write them as $(u_{n_1},w_{n_1}),\dots ,(u_{n_m},w_{n_m})$. Since $(a+\delta ,b]\subseteq {\bigcup }_{j=1}^m(u_{n_j},w_{n_j}]$, monotonicity and finite additivity give

$F(b)-F(a+\delta )\le {\sum }_{j=1}^m(F(w_{n_j})-F(u_{n_j}))<{\sum }_{j=1}^m{\mu }_{0,F}(I_{n_j})+\frac{\epsilon }{2}\le {\sum }_n{\mu }_{0,F}(I_n)+\frac{\epsilon }{2}\mathrm{.}$

Adding $F(a+\delta )-F(a)<\epsilon \mathrm{/}2$ yields

$F(b)-F(a)<{\sum }_n{\mu }_{0,F}(I_n)+\epsilon \mathrm{.}$

[step 2.1, step 3.1, L1, algebra]

Together with step 2.1, this proves countable additivity for bounded intervals. [step 2.1, step 4.1]

By step 5.1,

$F(b)-F(M)={\sum }_n{\mu }_{0,F}(I_n\cap (M,b])\le {\sum }_n{\mu }_{0,F}(I_n)\mathrm{.}$

Taking the supremum over $M<b$ gives ${\mu }_{0,F}(E)\le {\sum }_n{\mu }_{0,F}(I_n)$. Combined with step 2.1, this proves countable additivity for left rays. The same argument with $(a,M]$ proves the right-ray case $E=(a,\mathrm{\infty })$. [step 2.1, step 5.1, algebra]

So step 5.1 gives

$F(N)-F(M)={\sum }_n{\mu }_{0,F}(I_n\cap (M,N])\le {\sum }_n{\mu }_{0,F}(I_n)\mathrm{.}$

Taking the supremum over $M<N$ yields ${\mu }_{0,F}(\mathbb{R})\le {\sum }_n{\mu }_{0,F}(I_n)$. Together with step 2.1 and step 6.1, this proves countable additivity for every h-interval. [step 2.1, step 5.1, step 6.1, algebra]

Therefore ${\mu }_{0,F}$ is a premeasure. [step 1.1, step 5.1, step 6.1, step 7.1] ∎

By [L3], that sigma-algebra is $\mathcal{B}(\mathbb{R})$, and therefore

$\mu ((a,b])={\mu }_{0,F}((a,b])=F(b)-F(a)$

for every $a<b$. [L1, L2, L3]

So monotonicity and step 1.1 give

$\mu (K)\le \mu ((-R-1,R])=F(R)-F(-R-1)<+\mathrm{\infty }\mathrm{.}$

Thus $\mu$ is finite on compact sets. [step 1.1, L4, algebra]

This is the required ${\mu }_F$. [step 1.1, step 2.1] ∎

$\mathcal{P}:=\{(a,b]:a<b\}\mathrm{.}$

If $I_1=(a,b]$ and $I_2=(c,d]$ lie in $\mathcal{P}$, then $I_1\cap I_2$ is either empty or another half-open interval $(\max \{a,c\},\min \{b,d\}]$, so $\mathcal{P}$ is a pi-system. By [L1], $\sigma (\mathcal{P})=\mathcal{B}(\mathbb{R})$. [L1, algebra]

If $0\le x<y$, then $(0,x]\subseteq (0,y]$, so $F_{\mu }(x)\le F_{\mu }(y)$. If $x<y\le 0$, then $(y,0]\subseteq (x,0]$, so $-\mu ((x,0])\le -\mu ((y,0])$, again giving $F_{\mu }(x)\le F_{\mu }(y)$. If $x<0\le y$, then $F_{\mu }(y)-F_{\mu }(x)=\mu ((0,y])+\mu ((x,0])=\mu ((x,y])\ge 0$. [given, algebra]

$0={\mu }_F((a,b])-{\mu }_G((a,b])=(F(b)-G(b))-(F(a)-G(a))\mathrm{.}$

So $F(b)-G(b)=F(a)-G(a)$ for all $a<b$, and therefore $F-G$ is constant. [algebra]

In the three sign cases:

$\mu ((0,b])-\mu ((0,a])=\mu ((a,b])(0\le a<b),$

$-\mu ((b,0])+\mu ((a,0])=\mu ((a,b])(a<b\le 0),$

and

$\mu ((0,b])+\mu ((a,0])=\mu ((a,b])(a<0\le b)\mathrm{.}$

So the displayed interval formula always holds. [step 1.1, given, algebra]

For all large $n$ one has $x+h_n<0$ when $x<0$, while for $x\ge 0$ no sign change occurs. In either case, step 2.1 gives

$F_{\mu }(x+h_n)-F_{\mu }(x)=\mu ((x,x+h_n])\mathrm{.}$

The sets $(x,x+h_n]$ decrease to $\mathrm{\varnothing }$, and the first one has finite measure because it is contained in a compact interval. Therefore [L1] gives $\mu ((x,x+h_n])\to 0$, so $F_{\mu }(x+h_n)\to F_{\mu }(x)$. [step 2.1, L1]

Step 2.1 says that ${\mu }_{F_{\mu }}$ and $\mu$ agree on every half-open interval, so [L4] gives ${\mu }_{F_{\mu }}=\mu$. [step 2.1, step 3.1, L3, L4]

The interval values of ${\mu }_F$ and ${\mu }_G$ therefore agree by [L3], and [L4] yields ${\mu }_F={\mu }_G$. Together with steps 1.1, 1.2, 1.3, 2.1, and 3.1 this proves the theorem. [step 1.1, step 1.2, step 1.3, step 2.1, step 3.1, L3, L4] ∎

Because $(a,b)={\bigcup }_n(a,t_n]$, continuity from below and [L1] give

${\mu }_F((a,b))={\lim }_n{\mu }_F((a,t_n])={\lim }_n(F(t_n)-F(a))=F(b^{-})-F(a)\mathrm{.}$

Here $F{\mathrm{\mid }}_{[a,b]}$ has bounded variation because it is nondecreasing, so [L3] ensures that the displayed left limit exists. [L1, L2, L3, algebra]

The first interval has finite measure because $F$ is real-valued, so continuity from above and [L1] give

${\mu }_F([a,b])={\lim }_n{\mu }_F((s_n,b])={\lim }_n(F(b)-F(s_n))=F(b)-F(a^{-})\mathrm{.}$

The restriction $F{\mathrm{\mid }}_{[a-1,a]}$ has bounded variation because it is nondecreasing, so [L3] ensures that the displayed left limit exists. [L1, L2, L3, algebra]

Indeed, $(s_0,a]$ has finite measure and ${\mu }_F(\{a\})={\lim }_n{\mu }_F((s_n,a])={\lim }_n(F(a)-F(s_n))=F(a)-F(a^{-})\mathrm{.}$ [step 2.1, L1, L2, L3]

The same argument gives ${\mu }_F(\{b\})=F(b)-F(b^{-})$. Therefore

${\mu }_F([a,b))={\mu }_F([a,b])-{\mu }_F(\{b\})=F(b^{-})-F(a^{-})\mathrm{.}$

Together with step 1.1, this proves all four interval formulas. [step 1.1, step 2.1, step 3.1, L1, L6, algebra]

So [L3] makes its discontinuity set at most countable. Every atom of ${\mu }_F$ in $(-m,m)$ is an interior jump point by step 4.1, hence lies in that countable discontinuity set. Therefore the atoms in $(-m,m)$ are at most countable. By [L4], their union over $m\ge 1$ is at most countable, and this union contains every atom of ${\mu }_F$. [given, step 4.1, L3, L4]

[step 1.1, step 2.1, step 3.1, step 4.1, step 5.1] ∎

Choose $\eta >0$ so small that $F((b+\eta {)}^{-})-F(b)<\epsilon$; then the open interval $U:=(a,b+\eta )$ contains $I$ and [L1] gives

${\mu }_F(U\setminus I)={\mu }_F((b,b+\eta ))<\epsilon \mathrm{.}$

Likewise choose $\delta >0$ with $F((a+\delta {)}^{-})-F(a)<\epsilon$; then the closed interval $K:=[a+\delta ,b]$ is compact by [L4], satisfies $K\subseteq I$, and [L1] gives

${\mu }_F(I\setminus K)={\mu }_F((a,a+\delta ))=F((a+\delta {)}^{-})-F(a)<\epsilon \mathrm{.}$

[L1, L4, choose]

$A\subseteq U,{\mu }_F((U\cap I_m)\setminus A)<\epsilon \mathrm{.}$

By step 1.1, every trace $H\cap I_m$ with $H\in \mathcal{H}$ lies in ${\mathcal{O}}_m$. Those traces form an algebra on $I_m$, because $\mathcal{H}$ is an algebra on $\mathbb{R}$ and traces preserve complements and finite unions. By [L5] together with the generator statement of [L3], they generate the Borel sigma-algebra of the subspace $I_m$. [step 1.1, L3, L5]

Then $U:={\bigcup }_n{U}_n$ is open, contains $A$, and

$(U\cap I_m)\setminus A\subseteq {\bigcup }_n((U_n\cap I_m)\setminus A_n),$

so ${\mu }_F((U\cap I_m)\setminus A)<\epsilon$. If instead $A_n\downarrow A$, then $A_0\subseteq I_m$ has finite measure by [L1], so [L2] gives ${\mu }_F(A_n\setminus A)\to 0$; choose $n$ with ${\mu }_F(A_n\setminus A)<\epsilon \mathrm{/}2$, then choose open $U_n\supseteq A_n$ with ${\mu }_F((U_n\cap I_m)\setminus A_n)<\epsilon \mathrm{/}2$. The open set $U_n$ contains $A$, and

${\mu }_F((U_n\cap I_m)\setminus A)\le {\mu }_F((U_n\cap I_m)\setminus A_n)+{\mu }_F(A_n\setminus A)<\epsilon \mathrm{.}$

[step 2.1, L1, L2]

Therefore [L3] gives that every Borel subset of $I_m$ lies in ${\mathcal{O}}_m$: every bounded Borel set is relatively outer regular inside a compact interval. [step 2.1, step 3.1, L3]

Choose open $U\supseteq I_m\setminus A$ with ${\mu }_F((U\cap I_m)\setminus (I_m\setminus A))<\epsilon$. Put $K:=I_m\setminus U$. Then $K$ is closed in the compact interval $I_m$, hence compact by [L4], and $K\subseteq A$. Moreover

$A\setminus K=A\cap U=(U\cap I_m)\setminus (I_m\setminus A),$

so ${\mu }_F(A\setminus K)<\epsilon$. Thus every bounded Borel set is also inner regular by compact sets. [step 4.1, L4]

Then $E_m\uparrow E$, so [L2] gives ${\mu }_F(E)={\sup }_m{\mu }_F(E_m)$. For each $m$, step 5.1 gives a compact $K_m\subseteq E_m\subseteq E$ with ${\mu }_F(E_m\setminus K_m)<1\mathrm{/}(m+1)$. Hence

$\sup \{{\mu }_F(K):K\subseteq E,K\text{ compact}\}\ge {\sup }_m{\mu }_F(K_m)={\sup }_m{\mu }_F(E_m)={\mu }_F(E),$

while the reverse inequality is monotonicity. [step 5.1, L2, algebra]

This is possible directly from [L2] on $E\cap I_m\uparrow E$. Step 4.1 gives an open $V\supseteq E\cap I_m$ with ${\mu }_F((V\cap I_m)\setminus E)<\epsilon \mathrm{/}3$. For each of the two tails $E\cap (m,\mathrm{\infty })$ and $E\cap (-\mathrm{\infty },-m)$, decompose into unit half-open strips and apply step 4.1 on each compact strip, choosing the errors summably below $\epsilon \mathrm{/}3$; enlarging each strip by a thin open shell whose ${\mu }_F$-measure is also chosen summably below $\epsilon \mathrm{/}3$ via [L1], the union of those stripwise open sets is open and contributes total excess below $2\epsilon \mathrm{/}3$. Taking the union with $V$ gives an open set $U\supseteq E$ with ${\mu }_F(U\setminus E)<\epsilon$. [step 4.1, step 6.1, L1, L2]

So $\inf \{{\mu }_F(U):E\subseteq U,U\text{ open}\}=+\mathrm{\infty }={\mu }_F(E)$. Combining this with step 7.1 gives the outer regularity equality in all cases. The inner regularity equality is step 6.1, and the finite-measure $\epsilon$-approximation statement is exactly the combination of steps 6.1 and 7.1. [step 6.1, step 7.1, algebra] ∎

${\mu }_{\mathrm{id}}((a,b])={\mathrm{id}}_{\mathbb{R}}(b)-{\mathrm{id}}_{\mathbb{R}}(a)=b-a\mathrm{.}$

[L1]

Therefore [L3] gives ${\mu }_{\mathrm{id}}=\lambda$. [step 1.1, step 1.2, L3] ∎