Measurable Densities and Radon Volume on Manifolds

For $n\ge 1$ put $W=x(U\cap V\cap \mathrm{Int}M)$ and $Z=y(U\cap V\cap \mathrm{Int}M)$. These are open in ${\mathbb{R}}^n$. Boundary invariance makes $T=y\circ x^{-1}:W\to Z$ a smooth diffeomorphism. The images of $E\cap \mathrm{Int}M$ are Borel: charts are homeomorphisms and the trace sigma-algebra agrees with relative Borel sets.

On $Z$ take $h={\mathbf{1}}_{y(E\cap \mathrm{Int}M)}{r}_y$, with the zero-times-infinity convention. It is nonnegative Borel. For $u\in W$, $h(Tu)\mathrm{\mid }\det DT(u)\mathrm{\mid }={\mathbf{1}}_{x(E\cap \mathrm{Int}M)}(u)r_x(u)$. Applying the Borel substitution formula to these exact domains and this integrand equates the two interior integrals.

The remaining coordinate pieces of $E$ lie in the face $u_n=0$, or are empty for an interior chart. Their integrals vanish by nullity, without boundedness of $r_x$ or $r_y$. Adding them back proves the formula for $n\ge 1$.

For $n=0$ each nonempty chart has one point. Thus $E$ is empty or the common singleton. Both integrals are respectively zero or the same weight, because the transition determinant is one. This proves the formula in every dimension, including zero or infinite weight.

For each chart set $q_i=({\phi }_i\circ x_i^{-1})r_{x_i}$. This is nonnegative Borel. The set function ${\nu }_i(E)={\int }_{x_i(E\cap U_i)}{q}_{i}d{\lambda }_n$ is a measure: disjoint Borel sets have disjoint Borel chart images, and the indefinite-integral theorem supplies countable additivity there. In dimension zero it is a singleton weight times its indicator, hence also a measure, even for infinite weight.

For disjoint Borel $E_k$, ${\nu }_i({\bigcup }_k{E}_k)={\sum }_k{\nu }_i(E_k)$; equivalently apply the nonnegative summation theorem to $q_i{\mathbf{1}}_{x_i(E_k\cap U_i)}$. For $a_{ik}={\nu }_i(E_k)\ge 0$, both iterated sums equal ${\sup }_{m,l}{\sum }_{i\le m,k\le l}{a}_{ik}$: a finite selection in any row fits in some finite rectangle, and conversely every rectangle is bounded by either iterated sum. Consequently $\mu ({\bigcup }_k{E}_k)={\sum }_k\mu (E_k)$.

Every ${\nu }_i(\mathrm{\varnothing })=0$, so $\mu (\mathrm{\varnothing })=0$; all values are nonnegative extended reals. Zero coefficients contribute zero, and a one-term family gives its chart measure. Thus the claimed Borel measure exists with no finiteness assumption.

For a finite nonnegative Borel scalar $a$ on $M$, $ar$ is again a density: in coordinates its coefficient is $(a\circ x^{-1})r_x$, and multiplying the transition law by $a\circ x^{-1}$ proves its law, also at $r_x=\mathrm{\infty }$ when $a=0$. Thus the overlap lemma applies to ${\phi }_{i}r$ and ${\phi }_i{\psi }_{j}r$.

Fix a Borel $E\subset U$. Transfer the $i$th gluing term, for the density ${\phi }_{i}r$, from $x_i$ to $x$ on $E\cap U_i$. In the $x$ chart the transferred integrands sum to $r_x{\mathbf{1}}_{x(E)}$: if $r_x$ is finite this is distributivity and ${\sum }_i{\phi }_i=1$; if $r_x=\mathrm{\infty }$, at least one of the finitely many nonzero weights is positive, so the sum is infinite. The nonnegative summation theorem therefore gives the chart-restriction identity.

For two partitions $({\phi }_i)$ and $({\psi }_j)$, expand the first construction on a Borel $E$ as ${\sum }_i{\sum }_j{\int }_{x_i(E\cap U_i\cap V_j)}({\phi }_i{\psi }_{j}r{)}_{x_i}$. This expansion follows from the same pointwise argument, now summing ${\psi }_j$, and the nonnegative summation theorem. The overlap lemma transfers each summand to $y_j$. Both iterated sums are the supremum of the finite rectangular subsums, so reordering and summing ${\phi }_i$ gives the second construction.

Thus the two Borel measures are identical on every Borel set. All formulas remain true for the empty set and zero density, giving zero; in dimension zero the chart formula is the singleton weight. Neither infinity nor boundary points require cancellation: the overlap lemma already treats both.

For each point $p$ in positive dimension, a chart contains a relative closed ball or half-ball $H$ around its coordinate image. Choose it bounded with closure inside the chart image. Its inverse image $K$ is compact and contains a neighborhood $W$ of $p$. The continuous coefficient on $H$ is bounded by a finite $C$, so ${\mu }_r(W)\le {\mu }_r(K)={\int }_H{r}_x\le C{\lambda }_n(H)<\mathrm{\infty }$. For $n=0$ take $W=K=\{p\}$, whose measure is the finite coefficient $r(p)$.

The neighborhoods $W$ cover any compact $K_0$ finitely, giving ${\mu }_r(K_0)\le {\sum }_{j=1}^m{\mu }_r(W_j)<\mathrm{\infty }$. They also show local finiteness. To get a countable cover, take the members of a countable base that are contained in some such $W$; these cover $M$ and individually have finite measure. Enumerating these basis members proves sigma-finiteness without selecting neighborhoods at every point.

The same compact chart neighborhoods show local compactness also at the boundary; Hausdorffness and second countability are standing assumptions. The compact-finite Borel measure therefore satisfies the regularity theorem. Its conclusion includes the outer and open-set inner regularity required by the stated Radon convention.

Apply the completion theorem to $(M,\mathcal{B}(M),{\mu }_r)$ under the standing countable choice. Explicitly, $E=B\cup N$ with $B,Z$ Borel, $N\subset Z$ and ${\mu }_r(Z)=0$ has ${\bar{\mu }}_r(E)={\mu }_r(B)$. Empty $M$ and empty compact sets have mass zero; a singleton in dimension zero has its finite positive weight. No total-mass bound is asserted.

Choose a finite $C\ge 0$ with $\mathrm{\mid }f\mathrm{\mid }\le C$, and put $g=f+C\ge 0$. Its Borel measurability and $g\le 2C{\mathbf{1}}_Q$ imply ${\int }_{Q}g\le 2C{\prod }_j(b_j-a_j)<\mathrm{\infty }$. Likewise ${\int }_Q\mathrm{\mid }f\mathrm{\mid }\le C{\prod }_j(b_j-a_j)$, so $f$ is integrable.

For a finite grid $P$, list its closed cells $Q_1,\dots ,Q_m$ and disjointify them as $D_i=Q_i\setminus {\bigcup }_{j<i}{Q}_j$. These are Borel and partition $Q$; each $D_i$ contains the interior of $Q_i$ and differs from it only by grid faces. Thus ${\lambda }_n(D_i)=\mathrm{vol}(Q_i)$. Let $m_i={\inf }_{Q_i}g$ and $M_i={\sup }_{Q_i}g$. Each cell is nonempty and $g$ bounded, so these are finite.

The nonnegative simple functions $l_P={\sum }_i{m}_i{\mathbf{1}}_{D_i}$ and $u_P={\sum }_i{M}_i{\mathbf{1}}_{D_i}$ satisfy $l_P\le g\le u_P$ everywhere, including every assigned face. Their integrals are respectively $L(g,P)$ and $U(g,P)$. Hence $L(g,P)\le {\int }_{Q}g\le U(g,P)$.

Every Riemann sum of $g$ equals the corresponding sum of $f$ plus $C\mathrm{vol}(Q)$, so $g$ is Riemann integrable with value $J=I_R(f)+C\mathrm{vol}(Q)$. Darboux equivalence gives ${\sup }_{P}L(g,P)={\inf }_{P}U(g,P)=J$. Taking supremum and infimum in the preceding bounds yields ${\int }_{Q}g=J$.

The constant $C$ is integrable on $Q$, so linearity gives ${\int }_{Q}f={\int }_{Q}g-C{\lambda }_n(Q)=I_R(f)$. If $C=0$, all these quantities are zero; the proof also applies to $n=1$ and constant-one integrands. Degenerate boxes and dimension zero are outside the stated domain.

For $f={\mathbf{1}}_E$, with $E$ Borel, the left side is ${\mu }_r(E)$ and the right side is its defining chart sum. For a nonnegative simple function $s={\sum }_{k=1}^m{a}_k{\mathbf{1}}_{E_k}$ on disjoint Borel sets, finite additivity and homogeneity of the nonnegative integral give the formula term by term; a finite sum interchanges with the nonnegative chart sum. Coefficients $a_k=0$ contribute zero even when ${\mu }_r(E_k)=\mathrm{\infty }$.

Take nonnegative simple Borel $s_m\uparrow f$. For each chart, $s_m(x_i^{-1}(u))q_i(u)\uparrow f(x_i^{-1}(u))q_i(u)$, where $q_i=({\phi }_i\circ x_i^{-1})r_{x_i}$. This also holds if $q_i=\mathrm{\infty }$: a positive limiting value of $f$ forces an eventually positive $s_m$, whereas a zero limit makes every $s_m$ zero. Monotone convergence on $M$ and each chart, followed by ${\sup }_m{\sum }_i{a}_{mi}={\sum }_i{\sup }_m{a}_{mi}$ for increasing nonnegative $a_{mi}$, proves the formula. The last identity follows by taking finite chart sums first and then their supremum.

For real or complex Borel $f$ with $\int \mathrm{\mid }f\mathrm{\mid }d{\mu }_r<\mathrm{\infty }$, apply the nonnegative formula to $\mathrm{\mid }f\mathrm{\mid }$. It gives ${\sum }_i\int \mathrm{\mid }f\circ x_i^{-1}\mathrm{\mid }{q}_i<\mathrm{\infty }$, so every chart term is integrable and the sum of their absolute integrals is finite. Apply the formula separately to $f^{+},f^{-}$, or to the four positive/negative real/imaginary parts. Subtraction now involves only finite numbers and produces the asserted absolutely convergent series. Where $q_i=\mathrm{\infty }$ and $f\mathrm{\ne }0$ the weighted absolute integrand $h_i$ is infinite only on a Lebesgue-null set: for $E_i=\{h_i=\mathrm{\infty }\}$ and every integer $m\ge 1$, $m{\lambda }_n(E_i)\le \int h_i<\mathrm{\infty }$, forcing ${\lambda }_n(E_i)=0$; one may set the signed coefficient to zero there. This leaves each part integral unchanged.

Let $f\ge 0$ be measurable for the completed measure. Choose increasing completed-simple $s_m\uparrow f$. For each of the countably many level sets in these simple functions, the completion definition supplies a Borel replacement with symmetric difference contained in a Borel null set. Countable choice selects these replacements; their exceptional Borel sets have a null union $N$. The replacement simple functions $t_m$ are nonnegative Borel and equal $s_m$ off $N$. Put $u_m={\max }_{k\le m}{t}_k$ off $N$ and zero on $N$. Then $u_m$ is increasing Borel, and $g={\sup }_m{u}_m$ is Borel and equals $f$ off $N$. Applying monotone convergence to $u_m$, whose Borel and completed integrals agree by the extension property, gives equality of the Borel and completed integrals of $g$. The null-set integral property gives $\int fd{\bar{\mu }}_r=\int gd{\mu }_r$. Thus the chart formula for $g$ computes the completed integral, including infinity.

Now let $r$ be positive smooth and $f$ smooth with compact support $K$. Local finiteness of the partition supports gives a finite subfamily meeting $K$: finitely many neighborhoods witnessing local finiteness cover $K$. Each coefficient of ${\phi }_{i}fr$ is smooth with compact support inside its chart. Its zero extension is bounded and Riemann integrable by the chart extension result; it is Borel because it is smooth on a Borel chart image and zero elsewhere. A bounding nondegenerate box and the local Darboux bridge identify its Riemann and Lebesgue integrals. There are only finitely many terms; their absolute integrals are finite by boundedness and bounded support. The nonnegative formula of step 2.1 applied to $\mathrm{\mid }f\mathrm{\mid }$ therefore shows $f$ is integrable, so step 3.1 applies and the sum equals the defining smooth integral.

In dimension zero, compact sets are finite: the singleton open cover has a finite subcover. The preceding smooth comparison is then the same finite sum ${\sum }_{p\in K}f(p)r(p)$ in both definitions. Empty $K$, the zero function, and empty $M$ all give zero. A singleton of weight one integrates $f$ to its value.

For completed real or complex $L^1$ functions apply the same construction to each nonnegative component and subtract, redefining on the exceptional Borel null set to get a finite-valued Borel representative. Its absolute integral is unchanged, so step 3.1 applies. Two Borel representatives differ inside a Borel null set $N$; the nonnegative chart formula for ${\mathbf{1}}_N$ makes each chart contribution of $N$ zero. Null-set invariance, and for $L^1$ also almost-everywhere equality, show independence of every representative choice.

Choose $p\in O$ and a chart around it restricted inside $O$. For $n\ge 1$, continuity and $r_x(x(p))>0$ give a relative coordinate neighborhood where $r_x\ge c=r_x(x(p))\mathrm{/}2>0$. This neighborhood contains a closed nondegenerate Euclidean box $Q$ in the interior of the half-space: even if $p$ is on the face, move its last coordinate a sufficiently small positive distance and choose a still smaller box. Hence ${\mu }_r(O)\ge {\int }_Q{r}_x\ge c\mathrm{vol}(Q)>0$.

For $n=0$ the singleton $\{p\}$ is open with measure $r(p)>0$, so again ${\mu }_r(O)>0$. The empty manifold has no nonempty open subset, making this clause vacuous.

If $q\in B_d(p,R)$, then $ϵ=R-d(p,q)>0$ and the triangle inequality gives $B_d(q,ϵ)\subset B_d(p,R)$. Thus the ball is open in the metric topology, hence in the manifold topology and Borel. It contains $p$ since $d(p,p)=0<R$, so the preceding positivity applies. If $\overline{B_d(p,R)}$ is compact, monotonicity and compact-finiteness give ${\mu }_r(B_d(p,R))\le {\mu }_r(\overline{B_d(p,R)})<\mathrm{\infty }$.

Take $M=\mathbb{R}$ with its identity chart and $r=\mathrm{\mid }dx\mathrm{\mid }$, coefficient one. This is a finite-valued positive smooth density, so its measure is locally finite and finite on compact sets. For every integer $N\ge 1$, ${\mu }_r([-N,N])={\int }_{-N}^{N}1dx=2N$.

For any finite proposed bound $L$, choose an integer $N>L\mathrm{/}2$. Monotonicity gives ${\mu }_r(\mathbb{R})\ge 2N>L$, so the total mass is infinite. The empty set and singleton have mass zero by the same length formula, and intervals of length one have mass one; none of these local values bounds the total.

Let $T(s,t)=(s+1,-t)$ act on $X=\mathbb{R}\times (-1,1)$ and let $q:X\to M=X\mathrm{/}⟨T⟩$ be the quotient. It is open since $q^{-1}q(V)={\bigcup }_{k\in \mathbb{Z}}{T}^{k}V$ is open for every open $V$. Rectangles of $s$-width less than one have disjoint translates, so $q$ restricts to a homeomorphism from each such rectangle onto an open chart. The quotient is Hausdorff: for two inequivalent points, first choose bounded neighborhoods; only finitely many integer translates can intersect them because their $s$-coordinates are bounded, and shrink the neighborhoods to exclude each of these finitely many intersections. Their quotient images then separate the two orbits. Images of rational rectangles form a countable base. Transition maps are restrictions of powers of $T$, hence smooth. If M had an orientation, its pullback to X would be a smooth sign $\sigma :X\to \{+1,-1\}$ relative to the coordinate frame. This sign is constant because X is connected (any two points are joined by a straight segment). But $DT=\mathrm{diag}(1,-1)$ implies $\sigma (Tp)=-\sigma (p)$, contradicting constancy. Thus M is a smooth nonorientable surface.

The explicit local coefficient one gives $\mathrm{\mid }dsdt\mathrm{\mid }$. Since $DT=\mathrm{diag}(1,-1)$ has absolute determinant one, all transition powers preserve this density. It descends to a positive smooth density on $M$, an explicit instance of the general existence theorem. Its coefficients are finite, so it defines a Radon measure. This nonorientable witness refutes the assertion.