Measurable Densities and Radon Volume on Manifolds: Examples

Take the identity chart on ${\mathbb{R}}^n$ and partition $\phi =1$. The coefficient is one, hence for each Borel $E$, ${\mu }_r(E)={\int }_{E}1d{\lambda }_n={\lambda }_n(E)$. In particular ${\mu }_r({\prod }_j[a_j,b_j])={\prod }_j(b_j-a_j)$; for the unit cube the result is one, and if a side has length zero the result is zero.

The equality on Borel sets identifies the completed domain and measure with those in the Lebesgue completion theorem. Thus the completion is $({\mathbb{R}}^n,\mathcal{L}({\mathbb{R}}^n),{\lambda }_n)$. Empty sets have zero measure in both domains; the case $n=1$ is the usual interval-length formula.

The function $x\mapsto 1\mathrm{/}x$ is positive, finite and smooth on $(0,1)$. On $[a,b]\subset (0,1)$ it is continuous and bounded; the logarithmic integral identity gives its Riemann integral $\log b-\log a=\log (b\mathrm{/}a)$. The Borel Darboux bridge equates the Lebesgue integral to this value. Removing the two null endpoints does not change it, so the chart formula gives ${\mu }_r((a,b))=\log (b\mathrm{/}a)$. For example $(a,b)=(1\mathrm{/}4,3\mathrm{/}4)$ gives $\log 3$.

Every compact subset of $(0,1)$ has finite measure by the positive smooth density theorem. More explicitly, it lies in $[a,b]\subset (0,1)$ and is bounded in measure by $\log (b\mathrm{/}a)$. The increasing sets $K_N=[1\mathrm{/}N,1-1\mathrm{/}N]$, $N\ge 3$, exhaust $(0,1)$ and have mass $\log (N-1)$. Monotone convergence applied to their indicators gives ${\mu }_r((0,1))={\lim }_N\log (N-1)=\mathrm{\infty }$. This divergence also follows without any limit identity for log: each interval $[2^j,2^{j+1}]$ contributes at least $1\mathrm{/}2$ to ${\int }_1^{2^m}dt\mathrm{/}t$, so these logarithms are unbounded.

Empty intervals and singletons have zero mass by the null-set formula. There is no endpoint value of the density at zero or one because neither belongs to the manifold; its blowup at the omitted zero endpoint is consistent with local finiteness.

Write $q:\mathbb{R}\to S^1$ for the quotient. The chart domains are $U_1=S^1\setminus \{q(\pi )\}$ and $U_2=S^1\setminus \{q(0)\}$. On one overlap component $x_2=x_1$, and on the other $x_2=x_1+2\pi$. Both derivatives equal one, so the two constant coefficients satisfy the density law.

The disjoint decomposition is $S^1=q((0,\pi ))\bigsqcup q((\pi ,2\pi ))\bigsqcup \{q(0),q(\pi )\}$. The chart formula assigns the two arcs masses $\pi$ and $\pi$, and the endpoints mass zero. Hence ${\mu }_r(S^1)=2\pi$.

For any subordinate pair $({\phi }_1,{\phi }_2)$, translate the negative-angle part of the first chart by $2\pi$. Translation has unit Jacobian. The chart-sum formula becomes ${\int }_0^{2\pi }({\phi }_1(q(t))+{\phi }_2(q(t)))dt={\int }_0^{2\pi }1dt=2\pi$, ignoring only the already null endpoints. Thus overlapping chart weights count each arc once. Empty arcs contribute zero.

The quoted strip construction gives a Hausdorff second-countable smooth nonorientable surface. Its chart transitions are powers of $T$, with derivative $\mathrm{diag}(1,(-1{)}^k)$. Thus $(ds{)}^2+((-1{)}^{k}dt{)}^2=d{s}^2+d{t}^2$ and $\mathrm{\mid }\det D{T}^k\mathrm{\mid }=1$; both the quadratic form and the unit density agree on overlaps and descend smoothly.

In such a chart an orthonormal frame has column matrix $A$ satisfying $A^{\mathsf{T}}A=I$. Taking determinants gives $(\det A{)}^2=1$, so the density on that frame is $\mathrm{\mid }\det A\mathrm{\mid }=1$. Conversely a density with this normalization must have coefficient one on the coordinate frame, which is orthonormal. This verifies the claimed normalization directly without a general Riemannian volume theorem. The positive finite coefficient also gives a Radon measure.

Let $q$ be the quotient map. The set $W=q((0,1)\times (-1,1))$ is a chart: no two points in this open strip are related by a nonzero power of $T$. Its mass is $1\cdot 2=2$. Its complement is the seam $S=q(\{0\}\times (-1,1))$, a Borel set because $W$ is open. In the seam chart $q((-1\mathrm{/}4,1\mathrm{/}4)\times (-1,1))$, $S$ is the coordinate line $s=0$; it has measure zero (or cover it by rectangles of arbitrarily small width). Hence ${\mu }_r(M)={\mu }_r(W)+{\mu }_r(S)=2$. The omitted edges $t=\pm 1$ are not points of $M$.

Each point has an open singleton chart, so M is discrete and every subset is Borel. Fix a countable base $(B_j)$. For each $p$ the base contains $\{p\}$; assigning the least such index injects M into $\mathbb{N}$. Thus M is countable without selecting a chart for each point. Singleton indicators form a locally finite smooth partition in dimension zero.

The singleton chart integral is ${\mu }_w(\{p\})=w(p){\lambda }_0({\mathbb{R}}^0)=w(p)$. Countable additivity on the disjoint singleton decomposition of any A gives ${\mu }_w(A)={\sum }_{p\in A}w(p)$. The empty sum is zero; zero or infinite weights cause no cancellation.

A compact subset of a discrete space is finite, since its singleton cover has a finite subcover. Every scalar function here is smooth in local zero-dimensional coordinates. Thus finite positive weights satisfy the positive smooth density theorem and give a Radon measure; directly, compact masses are finite sums, and every set is open and approximated in measure by its finite subsets.

For $M=\mathbb{N}$ and $w(k)=2^{-k-1}$, the partial sum through $k=N-1$ is $1-2^{-N}$, which tends to one. For $w(k)=1$ the same partial sum is N, hence the total mass is infinite. In the one-point case the formula gives precisely its weight.

On $\mathbb{R}$ the coefficient $r_x=1$ is positive and smooth. Its measure is locally finite, and ${\mu }_r([-N,N])={\int }_{-N}^{N}1dx=2N$ for every integer $N\ge 1$. Each compact K is bounded, hence contained in some $[-N,N]$ and has finite measure.

Given any finite $L>0$, an integer $N>L\mathrm{/}2$ yields ${\mu }_r(\mathbb{R})\ge 2N>L$, proving infinite total mass. Thus the hypothesis holds and the asserted finite-total-mass conclusion fails. The same chart computation gives ${\mu }_r(\mathrm{\varnothing })={\mu }_r(\{0\})=0$ and ${\mu }_r([0,1])=1$.

The inequality $\mathrm{\mid }x-1\mathrm{/}2\mathrm{\mid }<1\mathrm{/}4$ with $x\in (0,1)$ is equivalent to $1\mathrm{/}4<x<3\mathrm{/}4$. The closure $[1\mathrm{/}4,3\mathrm{/}4]$ is compact inside M, and the weighted interval computation gives $\mu (B(1\mathrm{/}2,1\mathrm{/}4))=\log ((3\mathrm{/}4)\mathrm{/}(1\mathrm{/}4))=\log 3$. The ball is open Borel and has positive finite measure.

Every $x\in (0,1)$ satisfies $\mathrm{\mid }x-1\mathrm{/}2\mathrm{\mid }<1\mathrm{/}2<1$, so $B(1\mathrm{/}2,1)=(0,1)$. Its mass is infinite by the interval example. Its closure in M is all of M, which is not compact: the open cover $\{(1\mathrm{/}N,1):N\ge 2\}$ of M has no finite subcover. Thus the finite-volume hypothesis on the closure is absent in precisely this example. Both radii are strictly positive; neither ball is empty.