Measurable integration extends smooth density integration

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.