Proof of Dirac Measures on Euclidean Space: Probability Measure, Integrals, Push-Forwards and the Coupling of Two Dirac Measures

Each result cited is universally quantified over the data in its own statement. Since $n$ and $x$ are arbitrary, Steps 1 and 2 below apply to the function of the preamble formed in any dimension at any point; Steps 4 and 5 use them in that generality. Membership in $\mathcal{P}(\mathbb{R}^{m})$ and integrals against such measures are those of Probability Measures on Euclidean Space and Random Vectors: Standing Notation §measures.

Step 1 (Clause 1: $\delta_{x}$ is a probability measure). The values of $\delta_{x}$ lie in $\{0,1\}\subseteq[0,\infty]$, and $\delta_{x}(\varnothing)=0$ because $x\notin\varnothing$. Let $(B_{m})_{m\in\mathbb{N}}$ be a sequence of pairwise disjoint members of $\mathcal{B}(\mathbb{R}^{n})$, let $U=\bigcup_{m\in\mathbb{N}}B_{m}$, and for $K\in\mathbb{N}$ let $S_{K}=\sum_{m=1}^{K}\delta_{x}(B_{m})$, a finite sum of real numbers. If $x\notin U$, then $x\notin B_{m}$ and $\delta_{x}(B_{m})=0$ for every $m$, so every $S_{K}$ equals its first summand $0$ by claim 7 of Properties of Finite Sums (single possibly nonzero summand, with index $1$); the partial sums are real, bounded above by $0$, with least upper bound $0$, so by the definition of the sum of a sequence in Measure, Measure Space, and Probability Measure, $\sum_{m}\delta_{x}(B_{m})=0=\delta_{x}(U)$. If $x\in U$, choose $j\in\mathbb{N}$ with $x\in B_{j}$. For $m\ne j$ the sets $B_{m}$ and $B_{j}$ are disjoint, so $x\notin B_{m}$ and $\delta_{x}(B_{m})=0$. Hence, by claim 7 of Properties of Finite Sums, $S_{K}=\delta_{x}(B_{j})=1$ for every $K$ with $j\le K$ (index $j$), and $S_{K}=\delta_{x}(B_{1})=0$ for every $K$ with $K<j$ (index $1$, all summands being $0$). Thus every partial sum is real and at most $1$, and $1=S_{j}$ is one of them, so $1$ is their least upper bound and, by the same definition, $\sum_{m}\delta_{x}(B_{m})=1=\delta_{x}(U)$. This is countable additivity, so $\delta_{x}$ is a measure on $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$ by Measure, Measure Space, and Probability Measure; since $x\in\mathbb{R}^{n}$ gives $\delta_{x}(\mathbb{R}^{n})=1$, it is a probability measure, that is, $\delta_{x}\in\mathcal{P}(\mathbb{R}^{n})$.

Step 2 (Clause 2). By Step 1, $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}),\delta_{x})$ is a measure space. By Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §finite-sets, $\{x\}\in\mathcal{B}(\mathbb{R}^{n})$, and $\delta_{x}(\{x\})=1$. Let $f:\mathbb{R}^{n}\to[0,\infty]$ be Borel, that is, measurable in the sense of Measure Spaces and the Lebesgue Integral: Standing Notation §measurable (see Probability Measures on Euclidean Space and Random Vectors: Standing Notation §borel-maps).

Lower bound. Let $t$ be a nonnegative real number with $t\le f(x)$. By The Integral of an Indicator Function is the Measure of the Set, $\mathbf{1}_{\{x\}}$ is a nonnegative simple function with $\int\mathbf{1}_{\{x\}}\,d\delta_{x}=\delta_{x}(\{x\})=1$, and by claim 1 of Linearity and Monotonicity of the Lebesgue Integral the map $t\mathbf{1}_{\{x\}}$ is measurable with $\int t\mathbf{1}_{\{x\}}\,d\delta_{x}=t$. Since $t\mathbf{1}_{\{x\}}(x)=t\le f(x)$ and $t\mathbf{1}_{\{x\}}(y)=0\le f(y)$ for $y\ne x$, the monotonicity in the same claim gives $t\le\int f\,d\delta_{x}$. If $f(x)=\infty$, every nonnegative real $t$ satisfies $t\le f(x)$ by the conventions of Measure, Measure Space, and Probability Measure; were $\int f\,d\delta_{x}$ a real number $c$, then $t=c+1$ would give $c+1\le c$, contradicting $c<c+1$ (claims 6 and 1 of Elementary Order Arithmetic in an Ordered Field); hence $\int f\,d\delta_{x}=\infty=f(x)$. If $f(x)$ is real, taking $t=f(x)$ gives $f(x)\le\int f\,d\delta_{x}$.

Upper bound when $c=f(x)$ is real. By the definition of measurability in Lebesgue Integral of a Nonnegative Measurable Function, $E=\{y\in\mathbb{R}^{n}:f(y)>c\}$ belongs to $\mathcal{B}(\mathbb{R}^{n})$; since $x\notin E$, $\delta_{x}(E)=0$, so $E$ is $\delta_{x}$-null by Null Set of a Measure. Every $y\notin E$ satisfies $f(y)\le c=c\mathbf{1}_{\mathbb{R}^{n}}(y)$, so $f\le c\mathbf{1}_{\mathbb{R}^{n}}$ holds $\delta_{x}$-almost everywhere (A Property Holding Almost Everywhere). The map $c\mathbf{1}_{\mathbb{R}^{n}}$ is measurable, and by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison, claim 1 of Linearity and Monotonicity of the Lebesgue Integral and The Integral of an Indicator Function is the Measure of the Set,

With the lower bound, $\int_{\mathbb{R}^{n}}f\,d\delta_{x}=f(x)$ in every case.

Real-valued $f$. Let $f:\mathbb{R}^{n}\to\mathbb{R}$ be Borel. Its positive and negative parts $f^{+},f^{-}$ of Integrable Function and the Lebesgue Integral are measurable, nonnegative and real-valued, and are integrated as $[0,\infty]$-valued maps (Probability Measures on Euclidean Space and Random Vectors: Standing Notation §measures). By the case just proved, $\int f^{+}\,d\delta_{x}=f^{+}(x)$ and $\int f^{-}\,d\delta_{x}=f^{-}(x)$, both finite. Hence $f$ is integrable by Integrable Function and the Lebesgue Integral, and $\int f\,d\delta_{x}=f^{+}(x)-f^{-}(x)=f(x)$, since $f=f^{+}-f^{-}$ by the same definition.

Step 3 (Clause 1: second moment). The map $y\mapsto\lVert y\rVert^{2}$ is a nonnegative Borel function by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §functions, so by the definition of the second moment and Step 2, $M_{2}(\delta_{x})=\lVert x\rVert^{2}$. This is a real number, hence $M_{2}(\delta_{x})<\infty$, and $\delta_{x}\in\mathcal{P}_{2}(\mathbb{R}^{n})$ by The Second Moment of a Probability Measure on Euclidean Space and the Probability Measures with Finite Second Moment §space.

Step 4 (Clause 3). Let $h:\mathbb{R}^{n}\to\mathbb{R}^{m}$ be Borel and $B\in\mathcal{B}(\mathbb{R}^{m})$. Then $h^{-1}(B)\in\mathcal{B}(\mathbb{R}^{n})$, and by Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward, $h_{\#}\delta_{x}(B)=\delta_{x}(h^{-1}(B))$. Since $x\in h^{-1}(B)$ exactly when $h(x)\in B$, this value is $1$ if $h(x)\in B$ and $0$ otherwise, which is $\delta_{h(x)}(B)$. The functions $h_{\#}\delta_{x}$ and $\delta_{h(x)}$ on $\mathcal{B}(\mathbb{R}^{m})$ therefore agree, so $h_{\#}\delta_{x}=\delta_{h(x)}$.

Step 5 (Clause 4). Let $x'\in\mathbb{R}^{n}$, put $z_{0}=\iota^{n,n}(x,x')\in\mathbb{R}^{n+n}$ and $\pi=\delta_{z_{0}}$. By Step 1 in dimension $n+n$, $\pi\in\mathcal{P}(\mathbb{R}^{n+n})$. By Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections, the projections $\mathrm{pr}_{1},\mathrm{pr}_{2}:\mathbb{R}^{n+n}\to\mathbb{R}^{n}$ are Borel with $\mathrm{pr}_{1}(z_{0})=x$ and $\mathrm{pr}_{2}(z_{0})=x'$. Step 4, applied in dimension $n+n$ with target dimension $n$ and $h=\mathrm{pr}_{1}$, respectively $h=\mathrm{pr}_{2}$, gives $(\mathrm{pr}_{1})_{\#}\pi=\delta_{x}$ and $(\mathrm{pr}_{2})_{\#}\pi=\delta_{x'}$; as $\delta_{x},\delta_{x'}\in\mathcal{P}(\mathbb{R}^{n})$ by Step 1, $\pi\in\Pi(\delta_{x},\delta_{x'})$ by Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost §coupling. The cost $I(\pi)$ of Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost §cost is the integral against $\pi$ of the nonnegative Borel function $z\mapsto\lVert\mathrm{pr}_{1}(z)-\mathrm{pr}_{2}(z)\rVert^{2}$ (Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §functions), so Step 2 in dimension $n+n$ gives

By Step 3, $\delta_{x},\delta_{x'}\in\mathcal{P}_{2}(\mathbb{R}^{n})$, so The Quadratic Wasserstein Distance on Euclidean Space §distance gives $W_{2}(\delta_{x},\delta_{x'})^{2}\le I(\pi)=\lVert x-x'\rVert^{2}$. Both $W_{2}(\delta_{x},\delta_{x'})$ and $\lVert x-x'\rVert$ are nonnegative, so claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field yields $W_{2}(\delta_{x},\delta_{x'})\le\lVert x-x'\rVert$.