TheoremBase

A bounded continuous function is cut off by a smooth bump and then mollified, which gives test functions bounded by the same constant and converging pointwise; dominated convergence then transfers equality of test-function integrals to bounded Lipschitz functions, and the positive and negative parts of a density are compared as two finite measures.

Proof

Throughout, each result cited is universally quantified over the data appearing in its own statement and is applied to the data named here. Let ι\iota denote the canonical map from N\mathbb{N} to R\mathbb{R}, positive with positive inverse by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field; this is the only use of the symbol ι\iota here. Closed balls Bˉ(x,r)\bar{B}(x,r) and the Lebesgue measure λd\lambda_{d} are those of Product Measures and Lebesgue Measure on ℝⁿ: Standing Notation and Background (ii) and Product Measures and Lebesgue Measure on ℝⁿ: Standing Notation and Background (v).

Step 1 (Claim 1). Let ff and MM be as in claim 1; we may assume 0≤M0\le M, since ∣f(x)∣≤M|f(x)|\le M forces it by claim 1 of Properties of the Absolute Value in an Ordered Field. Fix a mollifier kernel ρ\rho of radius 11 on Rd\mathbb{R}^{d}, which exists by Existence of Mollifier Kernels of Every Radius, and for a positive real ε\varepsilon let ρε(y)=(ε−1)dρ(ε−1y)\rho_{\varepsilon}(y)=(\varepsilon^{-1})^{d}\rho(\varepsilon^{-1}y), a mollifier kernel of radius ε\varepsilon by Rescaling a Mollifier Kernel.

Let n∈Nn\in\mathbb{N}. By Existence of Smooth Bump Functions on Euclidean Space, applied with x0=0Rdx_{0}=0_{\mathbb{R}^{d}}, r=ι(n)r=\iota(n) and s=ι(n)+1s=\iota(n)+1, there is a smooth map χn:Rd→R\chi_{n}:\mathbb{R}^{d}\to\mathbb{R} with 0≤χn≤10\le\chi_{n}\le1, with χn(x)=1\chi_{n}(x)=1 whenever ∥x∥≤ι(n)\lVert x\rVert\le\iota(n), and with χn(x)=0\chi_{n}(x)=0 whenever ι(n)+1≤∥x∥\iota(n)+1\le\lVert x\rVert. The product fχnf\chi_{n} is continuous on Rd\mathbb{R}^{d} by claim 3 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, χn\chi_{n} being continuous by claim 3 of Euclidean Space Is Open in Itself, and it vanishes outside Bˉ(0Rd,ι(n)+1)\bar{B}(0_{\mathbb{R}^{d}},\iota(n)+1).

Since fχnf\chi_{n} is continuous on the open set Rd\mathbb{R}^{d} and ρε\rho_{\varepsilon} is continuous and vanishes at every yy with ε<∥y∥\varepsilon<\lVert y\rVert, the convolution (fχn)∗ρε(f\chi_{n})*\rho_{\varepsilon} is defined on {x:Bˉ(x,ε)⊆Rd}=Rd\{x:\bar{B}(x,\varepsilon)\subseteq\mathbb{R}^{d}\}=\mathbb{R}^{d}, and it is smooth by Convolution with a CkC^k Kernel is of Class CkC^k. It is bounded by MM: with hx(y)=(fχn)(x−y)ρε(y)h_{x}(y)=(f\chi_{n})(x-y)\rho_{\varepsilon}(y) one has ∣hx∣≤Mρε|h_{x}|\le M\rho_{\varepsilon} pointwise, because ∣fχn∣≤M|f\chi_{n}|\le M and ρε≥0\rho_{\varepsilon}\ge0, so claims 1 and 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion and the unit mass of ρε\rho_{\varepsilon} give

∣((fχn)∗ρε)(x)∣≤M∫Rdρε dλd=M.\bigl|\bigl((f\chi_{n})*\rho_{\varepsilon}\bigr)(x)\bigr|\le M\int_{\mathbb{R}^{d}}\rho_{\varepsilon}\,d\lambda_{d}=M .

It also vanishes at every xx with ι(n)+1+ε<∥x∥\iota(n)+1+\varepsilon<\lVert x\rVert: for such xx and any yy with ρε(y)≠0\rho_{\varepsilon}(y)\ne0 one has ∥y∥≤ε\lVert y\rVert\le\varepsilon, hence ∥x−y∥≥∥x∥−∥y∥>ι(n)+1\lVert x-y\rVert\ge\lVert x\rVert-\lVert y\rVert>\iota(n)+1 by claim 6 of Properties of the Euclidean Norm on Rn\mathbb{R}^n: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, so hxh_{x} vanishes identically. Its support is therefore a closed subset of Bˉ(0Rd,ι(n)+1+ε)\bar{B}(0_{\mathbb{R}^{d}},\iota(n)+1+\varepsilon), which is compact by The Heine-Borel Theorem in Rn\mathbb{R}^n, and Compactness of Closed Balls and the Unit Sphere, hence compact by Closed Subset of a Compact Space is Compact; so (fχn)∗ρε(f\chi_{n})*\rho_{\varepsilon} is a test function.

The set Bˉ(0Rd,ι(n))\bar{B}(0_{\mathbb{R}^{d}},\iota(n)) is compact by The Heine-Borel Theorem in Rn\mathbb{R}^n, and Compactness of Closed Balls and the Unit Sphere, so claim 2 of Mollification Converges Uniformly on Compact Subsets, applied to fχnf\chi_{n} on Ω=Rd\Omega=\mathbb{R}^{d} with the kernel ρ\rho and the positive real number ι(n)−1\iota(n)^{-1}, provides a positive real εn\varepsilon_{n} with

∣((fχn)∗ρεn)(x)−(fχn)(x)∣<ι(n)−1for every x∈Bˉ(0Rd,ι(n)).\bigl|\bigl((f\chi_{n})*\rho_{\varepsilon_{n}}\bigr)(x)-(f\chi_{n})(x)\bigr|<\iota(n)^{-1}\qquad\text{for every }x\in\bar{B}(0_{\mathbb{R}^{d}},\iota(n)).

Put ψn=(fχn)∗ρεn\psi_{n}=(f\chi_{n})*\rho_{\varepsilon_{n}}. Then ψn∈Cc∞(Rd)\psi_{n}\in C_{c}^{\infty}(\mathbb{R}^{d}) and ∣ψn(x)∣≤M|\psi_{n}(x)|\le M for every xx, by the previous paragraph. If ∥x∥≤ι(n)\lVert x\rVert\le\iota(n) then χn(x)=1\chi_{n}(x)=1, so (fχn)(x)=f(x)(f\chi_{n})(x)=f(x) and ∣ψn(x)−f(x)∣<ι(n)−1|\psi_{n}(x)-f(x)|<\iota(n)^{-1}.

Finally, let x∈Rdx\in\mathbb{R}^{d}. By claim 1 of The Archimedean Property of the Real Numbers there is n0∈Nn_{0}\in\mathbb{N} with ∥x∥<ι(n0)\lVert x\rVert<\iota(n_{0}), and by claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field every n≥n0n\ge n_{0} satisfies ∥x∥<ι(n)\lVert x\rVert<\iota(n) and hence ∣ψn(x)−f(x)∣<ι(n)−1|\psi_{n}(x)-f(x)|<\iota(n)^{-1}. Since (ι(n)−1)n∈N(\iota(n)^{-1})_{n\in\mathbb{N}} converges to 00 by claim 3 of The Archimedean Property of the Real Numbers, the sequence (ψn(x))n∈N(\psi_{n}(x))_{n\in\mathbb{N}} converges to f(x)f(x). This proves claim 1.

Step 2 (Claim 2). Let μ,ν\mu,\nu be as in claim 2 and let f:Rd→Rf:\mathbb{R}^{d}\to\mathbb{R} be continuous and bounded, say ∣f∣≤M|f|\le M with 0≤M0\le M. Take (ψn)n∈N(\psi_{n})_{n\in\mathbb{N}} as in claim 1. The constant function with value MM is integrable with respect to μ\mu and with respect to ν\nu by claim 6 of Borel Measurability and Bounded Integration on a Metric Space, both measures being finite, and ∣ψn∣≤M|\psi_{n}|\le M for every nn while (ψn(x))n(\psi_{n}(x))_{n} converges to f(x)f(x) for every xx. Claim 7 of The Lebesgue Integral and Null Sets, applied on the measure space (Rd,B(Rd),μ)(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\mu) and then on (Rd,B(Rd),ν)(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\nu), gives that (∫ψn dμ)n\bigl(\int\psi_{n}\,d\mu\bigr)_{n} converges to ∫f dμ\int f\,d\mu and (∫ψn dν)n\bigl(\int\psi_{n}\,d\nu\bigr)_{n} converges to ∫f dν\int f\,d\nu. The two sequences are equal term by term by hypothesis, so their limits agree by Uniqueness of Limits in a Metric Space:

∫Rdf dμ=∫Rdf dν.\int_{\mathbb{R}^{d}}f\,d\mu=\int_{\mathbb{R}^{d}}f\,d\nu .

Every bounded Lipschitz f:Rd→Rf:\mathbb{R}^{d}\to\mathbb{R} is continuous by A Lipschitz Map is Uniformly Continuous and Continuous, so the displayed identity holds for all such ff, and claim 1 of Lipschitz Test Functions Determine a Finite Borel Measure, and Uniqueness of Weak Limits, applied to the metric space (Rd,dE)(\mathbb{R}^{d},d_{E}), gives μ(B)=ν(B)\mu(B)=\nu(B) for every B∈B(Rd)B\in\mathcal{B}(\mathbb{R}^{d}).

Step 3 (Claim 3). Let gg be as in claim 3 and put g+=max⁡(g,0)g^{+}=\max(g,0) and g−=max⁡(−g,0)g^{-}=\max(-g,0), Borel by claim 4 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps and nonnegative by claim 1 of Elementary Properties of the Maximum of Two Elements. At each point xx the identities g(x)=g+(x)−g−(x)g(x)=g^{+}(x)-g^{-}(x) and g+(x)+g−(x)=∣g(x)∣g^{+}(x)+g^{-}(x)=|g(x)| hold, by a case distinction on the sign of g(x)g(x): if 0≤g(x)0\le g(x) then g+(x)=g(x)g^{+}(x)=g(x) and g−(x)=0g^{-}(x)=0, while if g(x)<0g(x)<0 then g+(x)=0g^{+}(x)=0 and g−(x)=−g(x)g^{-}(x)=-g(x), in each case by claim 2 of Elementary Properties of the Maximum of Two Elements together with claim 1 of Properties of the Absolute Value in an Ordered Field. Both are integrable with respect to μ\mu, being nonnegative and dominated by the integrable ∣g∣|g|, by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion.

By claim 3 of Image Measures and Measures with a Density: Measures, Integrals and Change of Variables, applied on (Rd,B(Rd),μ)(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\mu) with the densities g+g^{+} and g−g^{-}, the set functions

μ+(A)=∫Rd1A g+ dμ,μ−(A)=∫Rd1A g− dμ(A∈B(Rd))\mu^{+}(A)=\int_{\mathbb{R}^{d}}\mathbf{1}_{A}\,g^{+}\,d\mu,\qquad \mu^{-}(A)=\int_{\mathbb{R}^{d}}\mathbf{1}_{A}\,g^{-}\,d\mu\qquad(A\in\mathcal{B}(\mathbb{R}^{d}))

are measures on (Rd,B(Rd))(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d})), hence Borel measures on (Rd,dE)(\mathbb{R}^{d},d_{E}), and for every AA-integrable datum the change-of-density formula of that clause holds. They are finite, since μ±(Rd)=∫g± dμ\mu^{\pm}(\mathbb{R}^{d})=\int g^{\pm}\,d\mu is a real number. Let ψ∈Cc∞(Rd)\psi\in C_{c}^{\infty}(\mathbb{R}^{d}). It is bounded, being continuous with compact support by A Continuous Compactly Supported Function on Rn\mathbb{R}^n is Bounded and Integrable, hence integrable with respect to μ+\mu^{+} and μ−\mu^{-} by claim 6 of Borel Measurability and Bounded Integration on a Metric Space, and claim 3 of Image Measures and Measures with a Density: Measures, Integrals and Change of Variables gives ∫ψ dμ±=∫ψ g± dμ\int\psi\,d\mu^{\pm}=\int\psi\,g^{\pm}\,d\mu. Therefore, by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion and the hypothesis,

∫Rdψ dμ+−∫Rdψ dμ−=∫Rdψ (g+−g−) dμ=∫Rdg ψ dμ=0.\int_{\mathbb{R}^{d}}\psi\,d\mu^{+}-\int_{\mathbb{R}^{d}}\psi\,d\mu^{-}=\int_{\mathbb{R}^{d}}\psi\,(g^{+}-g^{-})\,d\mu=\int_{\mathbb{R}^{d}}g\,\psi\,d\mu=0 .

So μ+\mu^{+} and μ−\mu^{-} integrate every test function alike, and claim 2 gives μ+=μ−\mu^{+}=\mu^{-}.

Let P={x∈Rd:0<g(x)}P=\{x\in\mathbb{R}^{d}:0<g(x)\}, a member of B(Rd)\mathcal{B}(\mathbb{R}^{d}) as the preimage of a ray under the Borel map gg. On PP one has g−=0g^{-}=0 and off PP one has g+=0g^{+}=0, by claim 1 of Elementary Properties of the Maximum of Two Elements. Hence 1Pg−\mathbf{1}_{P}g^{-} vanishes identically, so μ−(P)=0\mu^{-}(P)=0 by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, while 1Pg+=g+\mathbf{1}_{P}g^{+}=g^{+} everywhere, so μ+(P)=∫g+ dμ\mu^{+}(P)=\int g^{+}\,d\mu. From μ+(P)=μ−(P)\mu^{+}(P)=\mu^{-}(P) we get ∫Rdg+ dμ=0\int_{\mathbb{R}^{d}}g^{+}\,d\mu=0, and claim 4 of The Lebesgue Integral and Null Sets gives μ({x:g+(x)≠0})=0\mu(\{x:g^{+}(x)\ne0\})=0. The same argument applied to Q={x:g(x)<0}Q=\{x:g(x)<0\}, on which g+=0g^{+}=0 and off which g−=0g^{-}=0, gives μ({x:g−(x)≠0})=0\mu(\{x:g^{-}(x)\ne0\})=0.

Finally {x:g(x)≠0}={x:g+(x)≠0}∪{x:g−(x)≠0}\{x:g(x)\ne0\}=\{x:g^{+}(x)\ne0\}\cup\{x:g^{-}(x)\ne0\}, since g=g+−g−g=g^{+}-g^{-} and at each point at least one of g+,g−g^{+},g^{-} vanishes; it belongs to B(Rd)\mathcal{B}(\mathbb{R}^{d}), being the preimage of R∖{0}\mathbb{R}\setminus\{0\} under gg, and it is μ\mu-null by claim 1 of The Lebesgue Integral and Null Sets. This proves claim 3.

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