Proof of Integration by Parts on Euclidean Space Against a Compactly Supported Function of Class $C^{1}$

Each result cited is universally quantified over the data in its own statement. Throughout, $\mathcal{B}(\mathbb{R}^{q})$ is the $\sigma$-algebra $\mathcal{B}_{q}$ of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$ by claim 5 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets, and $\lambda_{q}$ is the measure $\lambda_{q}$ built there, by Lebesgue Measure on $\mathbb{R}^n$; $\lambda$ denotes Lebesgue measure on $\mathcal{B}(\mathbb{R})$ as in Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$, so that $\lambda=\lambda_{1}$; measurability of real-valued functions is that of Measurable Map, and the Set of Measurable Maps, which for real-valued functions agrees with the reading of Nonnegative Measurable Maps and Their Lebesgue Integral, as recorded in that definition. The set $\mathbb{R}^{q}$ is open by claim 1 of Euclidean Space Is Open in Itself. Points of $\mathbb{R}^{q}$ are tuples, and the coordinates of sums, differences and scalar multiples are the sums, differences and scalar multiples of the coordinates (Sum of Points of $\mathbb{R}^n$, Scalar Multiple of a Point of $\mathbb{R}^n$, claims 2 and 3 of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space).

Step 1: regularity of $\partial_{i}h$. By clause 1 of Functions of Class Cᵏ on an Open Subset of ℝⁿ (read through its clause 3), $h$ and $\partial_{i}h$ are continuous at every point of $\mathbb{R}^{q}$ in the sense of Continuity at a Point for Maps Between Euclidean Spaces, hence continuous on $\mathbb{R}^{q}$ in the sense of Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation (iv) by claim 1 of Euclidean Continuity Agrees with Metric Continuity for Real-Valued Functions. By claim 2 of Compact Support on $\mathbb{R}^n$ Means Vanishing Outside a Bounded Set there is a real number $R>0$ with $h(x)=0$ whenever $\lVert x\rVert>R$. Put $W=\{x\in\mathbb{R}^{q}:\lVert x\rVert>R\}$. The set $W$ is open: if $x\in W$ and $y\in\mathbb{R}^{q}$ satisfies $\lVert y-x\rVert<\lVert x\rVert-R$, then $\lVert x\rVert\le\lVert y\rVert+\lVert x-y\rVert$ by claim 6 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square applied to $x=y+(x-y)$, so $\lVert y\rVert>R$; thus the open ball of centre $x$ and radius $\lVert x\rVert-R$ lies in $W$ (claim 2 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square identifies $d_{E}(x,y)$ with $\lVert x-y\rVert$), which is openness in the metric sense, and metric openness is Euclidean openness by Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$. The restriction $h|_{W}$ is the zero function on $W$, so $h|_{W}=0\cdot h|_{W}$. By claim 3 of Restriction of a Cᵏ Function to an Open Subset, $h|_{W}$ is of class $C^{1}$ on $W$, so its partial derivative with respect to the $i$th variable exists at every point of $W$, and by the scalar-multiple rule in claim 1 of Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions, $\partial_{i}(h|_{W})(x)=\partial_{i}(0\cdot h|_{W})(x)=0\cdot\partial_{i}(h|_{W})(x)=0$ for every $x\in W$. By claim 1 of Restriction of a Cᵏ Function to an Open Subset this gives $\partial_{i}h(x)=0$ for every $x\in W$. Hence $\partial_{i}h$ is compactly supported by claim 2 of Compact Support on $\mathbb{R}^n$ Means Vanishing Outside a Bounded Set, and by claim 2 of A Continuous Compactly Supported Function on $\mathbb{R}^n$ is Bounded and Integrable it is measurable with respect to $\mathcal{B}(\mathbb{R}^{q})$ and $\mathcal{B}(\mathbb{R})$ and integrable.

Step 2: the slices. For $q\ge2$ write $\mathbb{R}^{q}=\mathbb{R}^{q-1}\times\mathbb{R}$ as in Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$, and for $\theta'=(\theta_{1},\dots,\theta_{q-1})\in\mathbb{R}^{q-1}$ and $t\in\mathbb{R}$ let $\theta'[t]=(\theta_{1},\dots,\theta_{i-1},t,\theta_{i},\dots,\theta_{q-1})\in\mathbb{R}^{q}$, the point whose $i$th coordinate is $t$ and whose remaining coordinates are those of $\theta'$ in order. If $q=1$ the parameter $\theta'$ is absent: every clause below of the form ``for every $\theta'\in\mathbb{R}^{q-1}$'' is read with $\theta'$ deleted, $\theta'[t]$ is $t$, a pair $(\theta',y)$ is $y$, and $\mathcal{B}_{q-1}\otimes\mathcal{G}$ and $\lambda_{q-1}\otimes\delta$ are $\mathcal{G}$ and $\delta$, in accordance with the conventions for $l=1$ stated in Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$. Fix $\theta'$ and define $\gamma:\mathbb{R}\to\mathbb{R}$ by $\gamma(t)=h(\theta'[t])$.

(a) $\gamma$ is differentiable at every $t\in\mathbb{R}$ with $\gamma'(t)=\partial_{i}h(\theta'[t])$. Fix $t$ and put $a=\theta'[t]$, a point of the open set $\mathbb{R}^{q}$ (claim 1 of Euclidean Space Is Open in Itself); in the notation of Slice Function and the Partial Derivative, $a[s]=\theta'[s]$ for every $s\in\mathbb{R}$. By claim 1 there choose an admissible radius $\rho>0$, and let $I=\{s:t-\rho<s<t+\rho\}$. Since $\partial_{i}h(a)$ exists, claim 2 there says that the slice function $\gamma|_{I}$ is differentiable at $t$ with derivative $\partial_{i}h(a)$: by Derivative of a Real Function on an Interval, for every $\varepsilon>0$ there is $\delta>0$ such that $|(\gamma(t+u)-\gamma(t))/u-\partial_{i}h(a)|<\varepsilon$ whenever $0<|u|<\delta$ and $t+u\in I$. Given $\varepsilon$, let $\delta'$ be the least of $\delta$ and $\rho$ (claim 9 of Elementary Order Arithmetic in an Ordered Field); then $0<|u|<\delta'$ forces $t+u\in I$, so the same estimate holds for all $u$ with $0<|u|<\delta'$, which is differentiability of $\gamma$ at $t$ with derivative $\partial_{i}h(a)$ (every $t$ is an interior point of the interval $\mathbb{R}$ by Basic Facts about Intervals of the Real Line (i)).

(b) $\gamma'$ is continuous, and $\gamma$ and $\gamma'$ vanish for $|t|>R$. The map $\tau:\mathbb{R}\to\mathbb{R}^{q}$, $\tau(t)=\theta'[t]$, satisfies $d_{E}(\tau(t),\tau(s))=|t-s|$: the difference $\tau(t)-\tau(s)$ has $i$th coordinate $t-s$ and all other coordinates $0$, so by claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square and claim 7 of Properties of Finite Sums, $\lVert\tau(t)-\tau(s)\rVert^{2}=(t-s)^{2}=|t-s|^{2}$ (claim 1 of Nonnegativity of Squares in an Ordered Field), and $|t-s|\ge0$, so $\lVert\tau(t)-\tau(s)\rVert=|t-s|$ by the uniqueness in claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square. Hence $\tau$ is continuous from $(\mathbb{R},d_{\mathbb{R}})$ to $(\mathbb{R}^{q},d_{E})$ (take $\delta=\varepsilon$), where $d_{\mathbb{R}}$ is the metric of The Absolute Value Metric on the Real Numbers. By (a), $\gamma'=\partial_{i}h\circ\tau$, which is continuous on $\mathbb{R}$ by Step 1 and claim 3 of Semicontinuity and Continuity Under Composition with a Continuous Map. Moreover $|t|\le\lVert\theta'[t]\rVert$ by claim 4 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, so if $|t|>R$ then $\theta'[t]\in W$ and, by the choice of $R$ and Step 1, $\gamma(t)=h(\theta'[t])=0$ and $\gamma'(t)=\partial_{i}h(\theta'[t])=0$.

(c) $\int_{\mathbb{R}}\gamma'\,d\lambda=0$. Put $S=R+1$; then $0<S$ and $-S<0<S$ by claims 6, 3 and 4 of Elementary Order Arithmetic in an Ordered Field, so $J=[-S,S]$ is an interval of which every $x$ with $-S<x<S$ is an interior point, by Basic Facts about Intervals of the Real Line (v). Let $G=\gamma|_{J}$ and $f_{0}=\gamma'|_{J}$. By claim 2 of Restriction Stability of Continuity and of the Derivative, $G$ is differentiable at every $x$ with $-S<x<S$ with $G'(x)=\gamma'(x)=f_{0}(x)$; by The Increment Form of the Derivative, and Differentiability Implies Continuity $\gamma$ is continuous at every point of $\mathbb{R}$, so $G$ is continuous on $J$ by claim 1 of Restriction Stability of Continuity and of the Derivative, and so is $f_{0}$, by (b) and the same claim 1. By claim 1 of Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval, $f_{0}$ is Riemann integrable on $J$, and Fundamental Theorem of Calculus, Part II: Evaluating an Integral by an Antiderivative gives $\int_{-S}^{S}f_{0}(x)\,dx=G(S)-G(-S)=\gamma(S)-\gamma(-S)=0$, both values vanishing by (b) since $S>R$. By claim 3 of Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval, the zero extension $\tilde f_{0}$ of $f_{0}$ satisfies $\int_{\mathbb{R}}\tilde f_{0}\,d\lambda=\int_{-S}^{S}f_{0}(x)\,dx=0$, and $\tilde f_{0}$ is integrable by claim 2 there. But $\tilde f_{0}=\gamma'$ as functions on $\mathbb{R}$: they agree on $J$, and off $J$ we have $|t|>S>R$, where $\gamma'(t)=0$ by (b). Hence $\gamma'$ is integrable with respect to $\lambda$ and $\int_{\mathbb{R}}\gamma'\,d\lambda=0$; that is,

$$\int_{\mathbb{R}}\partial_{i}h(\theta'[t])\,d\lambda(t)=0\qquad\text{for every }\theta'\in\mathbb{R}^{q-1}.$$

Step 3: assembly. Let $(Y,\mathcal{G},\delta)$ be the one-point measure space with unit mass at a point $z$ (claim 3 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions); since $\delta(Y)=1$, the measure $\delta$ is finite and hence $\sigma$-finite in the sense of Measure, Measure Space, and Probability Measure, and $\lambda_{q}$ is $\sigma$-finite by claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$. Define $F:\mathbb{R}^{q}\times Y\to\mathbb{R}$ by $F(\theta,z)=\partial_{i}h(\theta)$. For every Borel set $B\subseteq\mathbb{R}$, $F^{-1}(B)=(\partial_{i}h)^{-1}(B)\times Y$ is a measurable rectangle (Step 1), hence lies in $\mathcal{B}_{q}\otimes\mathcal{G}$ by Measurable Rectangles and the Product σ-Algebra; so $F$ is measurable, and likewise the functions $(\theta,z)\mapsto|\partial_{i}h(\theta)|$, $(\partial_{i}h)^{+}(\theta)$, $(\partial_{i}h)^{-}(\theta)$, which are $|F|$, $F^{+}$, $F^{-}$ in the notation of Integrable Functions and Their Lebesgue Integral (the functions $|\partial_{i}h|$, $(\partial_{i}h)^{\pm}$ on $\mathbb{R}^{q}$ are measurable by claim 4 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps and by Integrable Functions and Their Lebesgue Integral). For each of $u=|\partial_{i}h|,(\partial_{i}h)^{+},(\partial_{i}h)^{-}$, Tonelli (Tonelli's and Fubini's Theorems) applied to $(\theta,z)\mapsto u(\theta)$ and claim 3 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions for the inner integral give

$$\int_{\mathbb{R}^{q}\times Y}u(\theta)\,d(\lambda_{q}\otimes\delta)(\theta,z)=\int_{\mathbb{R}^{q}}\Bigl(\int_{Y}u(\theta)\,d\delta(z)\Bigr)d\lambda_{q}(\theta)=\int_{\mathbb{R}^{q}}u\,d\lambda_{q}.$$

With $u=|\partial_{i}h|$ the right side is finite (Step 1), so $F$ is integrable with respect to $\lambda_{q}\otimes\delta$ by Integrable Functions and Their Lebesgue Integral; with $u=(\partial_{i}h)^{\pm}$ and the definition of the integral of an integrable function, $\int_{\mathbb{R}^{q}\times Y}F\,d(\lambda_{q}\otimes\delta)=\int_{\mathbb{R}^{q}}\partial_{i}h\,d\lambda_{q}$.

Now apply claim 4 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$ with $l=q$, the index $i$, and $(Y,\mathcal{G},\mu)=(Y,\mathcal{G},\delta)$. In the notation of its claim 2, $F(\Psi_{i}(t,(\theta',z)))=\partial_{i}h(\theta'[t])$ for all $t\in\mathbb{R}$ and $\theta'\in\mathbb{R}^{q-1}$. Claim 4 provides a set $N\in\mathcal{B}_{q-1}\otimes\mathcal{G}$ with $(\lambda_{q-1}\otimes\delta)(N)=0$ such that the function $\Phi$ on $\mathbb{R}^{q-1}\times Y$, equal to $\int_{\mathbb{R}}\partial_{i}h(\theta'[t])\,d\lambda(t)$ at $(\theta',z)\notin N$ and to $0$ on $N$, is integrable and satisfies $\int\Phi\,d(\lambda_{q-1}\otimes\delta)=\int_{\mathbb{R}^{q}\times Y}F\,d(\lambda_{q}\otimes\delta)$. By Step 2(c), $\Phi$ is the zero function; as $\Phi$ is integrable and $\Phi=0\cdot\Phi$, claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion gives $\int\Phi\,d(\lambda_{q-1}\otimes\delta)=0\cdot\int\Phi\,d(\lambda_{q-1}\otimes\delta)=0$. Hence $\int_{\mathbb{R}^{q}}\partial_{i}h\,d\lambda_{q}=0$, which completes the proof of claim 1.

Claim 2. By claim 3 of Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions, $fg$ is of class $C^{1}$ on $\mathbb{R}^{q}$; it vanishes wherever $g$ does, so it is compactly supported by claim 2 of Compact Support on $\mathbb{R}^n$ Means Vanishing Outside a Bounded Set (applied to $g$ and then to $fg$). By claim 1 of Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions, $\partial_{i}(fg)=\partial_{i}f\,g+f\,\partial_{i}g$ on $\mathbb{R}^{q}$. As in Step 1, the functions $f$, $g$, $\partial_{i}f$, $\partial_{i}g$ are continuous on $\mathbb{R}^{q}$, and $\partial_{i}g$ is compactly supported by claim 1 applied to $h=g$. Hence $(\partial_{i}f)\,g$ and $f\,\partial_{i}g$ are continuous on $\mathbb{R}^{q}$ by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, and compactly supported since they vanish wherever $g$, respectively $\partial_{i}g$, vanishes; so they are integrable by claim 2 of A Continuous Compactly Supported Function on $\mathbb{R}^n$ is Bounded and Integrable. By claim 1 applied to $h=fg$ and the linearity of the integral for integrable functions (claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion),

$$0=\int_{\mathbb{R}^{q}}\partial_{i}(fg)\,d\lambda_{q}=\int_{\mathbb{R}^{q}}(\partial_{i}f)\,g\,d\lambda_{q}+\int_{\mathbb{R}^{q}}f\,\partial_{i}g\,d\lambda_{q},$$

which is the asserted identity.