TheoremBase

Test Functions Approximate Bounded Continuous Functions Pointwise, Determine a Finite Borel Measure, and Detect a Vanishing Density

Every bounded continuous function on Euclidean space is a pointwise limit of test functions bounded by the same constant; consequently two finite Borel measures that integrate every test function alike are equal, and an integrable Borel function whose products with all test functions integrate to zero vanishes almost everywhere.

Statement

In the setting of Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation, let Cc∞(Rd)C_{c}^{\infty}(\mathbb{R}^{d}) be the set of test functions on Rd\mathbb{R}^{d}, let B(Rd)\mathcal{B}(\mathbb{R}^{d}) be the Borel σ\sigma-algebra of Rd\mathbb{R}^{d}, which is that of the metric space (Rd,dE)(\mathbb{R}^{d},d_{E}), and call a Borel measure μ\mu on (Rd,dE)(\mathbb{R}^{d},d_{E}) finite if μ(Rd)<∞\mu(\mathbb{R}^{d})<\infty. Continuous and bounded are as fixed in Probability Measures on Euclidean Space and Random Vectors: Standing Notation (ii) and Bounded Real-Valued Function on a Set, and Borel and integrable as in Probability Measures on Euclidean Space and Random Vectors: Standing Notation (ii) and Probability Measures on Euclidean Space and Random Vectors: Standing Notation (iii).

1. (Pointwise approximation of a bounded continuous function) Let f:Rd→Rf:\mathbb{R}^{d}\to\mathbb{R} be continuous on Rd\mathbb{R}^{d} and let MM be a real number with ∣f(x)∣≤M|f(x)|\le M for every x∈Rdx\in\mathbb{R}^{d}. Then there is a sequence (ψn)n∈N(\psi_{n})_{n\in\mathbb{N}} in Cc∞(Rd)C_{c}^{\infty}(\mathbb{R}^{d}) such that ∣ψn(x)∣≤M|\psi_{n}(x)|\le M for every n∈Nn\in\mathbb{N} and every x∈Rdx\in\mathbb{R}^{d}, and such that for every x∈Rdx\in\mathbb{R}^{d} the sequence (ψn(x))n∈N(\psi_{n}(x))_{n\in\mathbb{N}} converges to f(x)f(x).

2. (Determination of a finite Borel measure) Let μ\mu and ν\nu be finite Borel measures on (Rd,dE)(\mathbb{R}^{d},d_{E}) such that

∫Rdψ dμ=∫Rdψ dνfor every ψ∈Cc∞(Rd),\int_{\mathbb{R}^{d}}\psi\,d\mu=\int_{\mathbb{R}^{d}}\psi\,d\nu\qquad\text{for every }\psi\in C_{c}^{\infty}(\mathbb{R}^{d}),

the integrals existing because each ψ\psi is Borel and, being continuous with compact support, bounded, so that claim 6 of Borel Measurability and Bounded Integration on a Metric Space applies. Then μ(B)=ν(B)\mu(B)=\nu(B) for every B∈B(Rd)B\in\mathcal{B}(\mathbb{R}^{d}).

3. (A density detected by test functions) Let μ∈P(Rd)\mu\in\mathcal{P}(\mathbb{R}^{d}) and let g:Rd→Rg:\mathbb{R}^{d}\to\mathbb{R} be Borel and integrable with respect to μ\mu. For ψ∈Cc∞(Rd)\psi\in C_{c}^{\infty}(\mathbb{R}^{d}) the product gψg\psi is Borel and integrable with respect to μ\mu, being dominated in absolute value by a real multiple of ∣g∣|g|, since ψ\psi is bounded. Suppose that

∫Rdg ψ dμ=0for every ψ∈Cc∞(Rd).\int_{\mathbb{R}^{d}}g\,\psi\,d\mu=0\qquad\text{for every }\psi\in C_{c}^{\infty}(\mathbb{R}^{d}).

Then the set {x∈Rd:g(x)≠0}\{x\in\mathbb{R}^{d}:g(x)\ne0\}, which belongs to B(Rd)\mathcal{B}(\mathbb{R}^{d}), satisfies

μ({x∈Rd:g(x)≠0})=0.\mu\bigl(\{x\in\mathbb{R}^{d}:g(x)\ne0\}\bigr)=0 .

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