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

Throughout, each result cited is universally quantified over the data appearing in its own statement and is applied to the data named here. Let $\iota$ denote the canonical map from $\mathbb{N}$ to $\mathbb{R}$, 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 $\iota$ here. Closed balls $\bar{B}(x,r)$ and the Lebesgue measure $\lambda_{d}$ are those of Euclidean Space and Lebesgue Measure: Standing Notation §space and Euclidean Space and Lebesgue Measure: Standing Notation §measure.

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

Let $n\in\mathbb{N}$. By Existence of Smooth Bump Functions on Euclidean Space, applied with $x_{0}=0_{\mathbb{R}^{d}}$, $r=\iota(n)$ and $s=\iota(n)+1$, there is a smooth map $\chi_{n}:\mathbb{R}^{d}\to\mathbb{R}$ with $0\le\chi_{n}\le1$, with $\chi_{n}(x)=1$ whenever $\lVert x\rVert\le\iota(n)$, and with $\chi_{n}(x)=0$ whenever $\iota(n)+1\le\lVert x\rVert$. The product $f\chi_{n}$ is continuous on $\mathbb{R}^{d}$ by claim 3 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, $\chi_{n}$ being continuous by claim 3 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous, and it vanishes outside $\bar{B}(0_{\mathbb{R}^{d}},\iota(n)+1)$.

Since $f\chi_{n}$ is continuous on the open set $\mathbb{R}^{d}$ and $\rho_{\varepsilon}$ is continuous and vanishes at every $y$ with $\varepsilon<\lVert y\rVert$, the convolution $(f\chi_{n})*\rho_{\varepsilon}$ is defined on $\{x:\bar{B}(x,\varepsilon)\subseteq\mathbb{R}^{d}\}=\mathbb{R}^{d}$, and it is smooth by Convolution with a $C^k$ Kernel is of Class $C^k$. It is bounded by $M$: with $h_{x}(y)=(f\chi_{n})(x-y)\rho_{\varepsilon}(y)$ one has $|h_{x}|\le M\rho_{\varepsilon}$ pointwise, because $|f\chi_{n}|\le M$ and $\rho_{\varepsilon}\ge0$, so claims 1 and 2 of Linearity and Monotonicity of the Lebesgue Integral and the unit mass of $\rho_{\varepsilon}$ give

It also vanishes at every $x$ with $\iota(n)+1+\varepsilon<\lVert x\rVert$: for such $x$ and any $y$ with $\rho_{\varepsilon}(y)\ne0$ one has $\lVert y\rVert\le\varepsilon$, hence $\lVert x-y\rVert\ge\lVert x\rVert-\lVert y\rVert>\iota(n)+1$ by claim 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, so $h_{x}$ vanishes identically. Its support is therefore a closed subset of $\bar{B}(0_{\mathbb{R}^{d}},\iota(n)+1+\varepsilon)$, which is compact by Heine-Borel Theorem in $\mathbb{R}^n$, hence compact by Closed Subset of a Compact Space is Compact; so $(f\chi_{n})*\rho_{\varepsilon}$ is a test function.

The set $\bar{B}(0_{\mathbb{R}^{d}},\iota(n))$ is compact by Heine-Borel Theorem in $\mathbb{R}^n$, so claim 2 of Mollification Converges Uniformly on Compact Subsets, applied to $f\chi_{n}$ on $\Omega=\mathbb{R}^{d}$ with the kernel $\rho$ and the positive real number $\iota(n)^{-1}$, provides a positive real $\varepsilon_{n}$ with

Put $\psi_{n}=(f\chi_{n})*\rho_{\varepsilon_{n}}$. Then $\psi_{n}\in C_{c}^{\infty}(\mathbb{R}^{d})$ and $|\psi_{n}(x)|\le M$ for every $x$, by the previous paragraph. If $\lVert x\rVert\le\iota(n)$ then $\chi_{n}(x)=1$, so $(f\chi_{n})(x)=f(x)$ and $|\psi_{n}(x)-f(x)|<\iota(n)^{-1}$.

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

Step 2 (Claim 2). Let $\mu,\nu$ be as in claim 2 and let $f:\mathbb{R}^{d}\to\mathbb{R}$ be continuous and bounded, say $|f|\le M$ with $0\le M$. Take $(\psi_{n})_{n\in\mathbb{N}}$ as in claim 1. The constant function with value $M$ is integrable with respect to $\mu$ and with respect to $\nu$ by claim 6 of Borel Measurability and Bounded Integration on a Metric Space, both measures being finite, and $|\psi_{n}|\le M$ for every $n$ while $(\psi_{n}(x))_{n}$ converges to $f(x)$ for every $x$. Claim 7 of The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere, applied on the measure space $(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\mu)$ and then on $(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\nu)$, gives that $\bigl(\int\psi_{n}\,d\mu\bigr)_{n}$ converges to $\int f\,d\mu$ and $\bigl(\int\psi_{n}\,d\nu\bigr)_{n}$ converges to $\int f\,d\nu$. The two sequences are equal term by term by hypothesis, so their limits agree by Uniqueness of Limits in a Metric Space:

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

Step 3 (Claim 3). Let $g$ be as in claim 3 and put $g^{+}=\max(g,0)$ and $g^{-}=\max(-g,0)$, Borel by claim 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions and nonnegative by claim 1 of Elementary Properties of the Maximum of Two Elements. At each point $x$ the identities $g(x)=g^{+}(x)-g^{-}(x)$ and $g^{+}(x)+g^{-}(x)=|g(x)|$ hold, by a case distinction on the sign of $g(x)$: if $0\le g(x)$ then $g^{+}(x)=g(x)$ and $g^{-}(x)=0$, while if $g(x)<0$ then $g^{+}(x)=0$ and $g^{-}(x)=-g(x)$, 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 $\mu$, being nonnegative and dominated by the integrable $|g|$, by claim 1 of Linearity and Monotonicity of the Lebesgue Integral.

By claim 3 of Image Measures, Measures with Densities, and Change of Variables, applied on $(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\mu)$ with the densities $g^{+}$ and $g^{-}$, the set functions

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

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

Let $P=\{x\in\mathbb{R}^{d}:0<g(x)\}$, a member of $\mathcal{B}(\mathbb{R}^{d})$ as the preimage of a ray under the Borel map $g$. On $P$ one has $g^{-}=0$ and off $P$ one has $g^{+}=0$, by claim 1 of Elementary Properties of the Maximum of Two Elements. Hence $\mathbf{1}_{P}g^{-}$ vanishes identically, so $\mu^{-}(P)=0$ by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, while $\mathbf{1}_{P}g^{+}=g^{+}$ everywhere, so $\mu^{+}(P)=\int g^{+}\,d\mu$. From $\mu^{+}(P)=\mu^{-}(P)$ we get $\int_{\mathbb{R}^{d}}g^{+}\,d\mu=0$, and claim 4 of The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere gives $\mu(\{x:g^{+}(x)\ne0\})=0$. The same argument applied to $Q=\{x:g(x)<0\}$, on which $g^{+}=0$ and off which $g^{-}=0$, gives $\mu(\{x:g^{-}(x)\ne0\})=0$.

Finally $\{x:g(x)\ne0\}=\{x:g^{+}(x)\ne0\}\cup\{x:g^{-}(x)\ne0\}$, since $g=g^{+}-g^{-}$ and at each point at least one of $g^{+},g^{-}$ vanishes; it belongs to $\mathcal{B}(\mathbb{R}^{d})$, being the preimage of $\mathbb{R}\setminus\{0\}$ under $g$, and it is $\mu$-null by claim 1 of The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere. This proves claim 3.