Proof of The Gaussian Weight Defines a Probability Distribution

We keep the notation of the statement, use the $[0,\infty]$-valued measurability and integral of Nonnegative Measurable Maps and Their Lebesgue Integral, and use the vocabulary jointly Borel of Joint Distribution, Expectations, and Block Independence for Independent Random Variables for maps on $\mathbb{R}^2$ measurable with respect to $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$. From Basic Properties of the Exponential Function: $\exp$ is positive, strictly increasing, satisfies $\exp(s+t)=\exp(s)\exp(t)$ and $\exp'=\exp$, and from the defining series $\exp(t)\ge1+t\ge t$ for $t\ge0$; the natural logarithm $\ln$ is its inverse, so for $s>0$ and real $w$: $\exp(w)>s\iff w>\ln s$, and $\exp(-\ln s)=1/s$ by the product property. Note $g$ is continuous, hence Borel measurable, and $0<g\le1$.

Preliminary (affine maps). Let $T(x)=\alpha x+\beta$ with real $\alpha\neq0$ and $\beta$, and write $\lambda^{*}$ for the Lebesgue outer measure. Images of intervals under $T$ are intervals with lengths multiplied by $|\alpha|$, so interval covers of $E$ map to interval covers of $T(E)$ and $\lambda^{*}(T(E))\le|\alpha|\lambda^{*}(E)$; applying the same to the affine map $T^{-1}$ gives equality, $\lambda^{*}(T(E))=|\alpha|\lambda^{*}(E)$ for every $E\subseteq\mathbb{R}$. For Borel $B$ both $T^{-1}(B)$ and $T(B)=(T^{-1})^{-1}(B)$ are Borel, being preimages under continuous maps (preimages of the generating open sets are open), and on Borel sets $\lambda$ is the restriction of $\lambda^{*}$ (claim 3 of Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness); hence

$$\lambda(T^{-1}(B))=\lambda^{*}(T^{-1}(B))=|\alpha|^{-1}\lambda^{*}(B)=|\alpha|^{-1}\lambda(B),$$

and in the same way $\lambda(T(B))=|\alpha|\lambda(B)$. By the standard machine (indicators: $\int\mathbf{1}_B\circ T\,d\lambda=\lambda(T^{-1}(B))=|\alpha|^{-1}\int\mathbf{1}_B\,d\lambda$; nonnegative simple functions by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion; general $h$ by Monotone Convergence Theorem along the dyadic staircase approximations $\varphi_L\circ h$, where $\varphi_L(t)=\min\{L,2^{-L}\lfloor2^{L}t\rfloor\}$ for $t\ge0$ and $L\in\mathbb{N}$, nonnegative simple functions increasing pointwise to $h$, whose compositions with $T$ increase to $h\circ T$), for every nonnegative Borel measurable $h$,

$$\int_{\mathbb{R}}h(\alpha x+\beta)\,d\lambda(x)=\frac{1}{|\alpha|}\int_{\mathbb{R}}h\,d\lambda .$$

Claim 1. $\nu(\emptyset)=0$. For pairwise disjoint Borel $B_j$: $\mathbf{1}_{\bigcup_jB_j}\,g=\sum_j\mathbf{1}_{B_j}g$ pointwise, the partial sums are nondecreasing, finite sums pull out of the integral by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, and Monotone Convergence Theorem gives $\nu(\bigcup_jB_j)=\sum_j\nu(B_j)$. So $\nu$ is a measure.

Claim 2. For $|x|\ge2$ we have $x^{2}/2\ge|x|$, so $g(x)\le\exp(-|x|)$ by monotonicity of $\exp$. Moreover $\int_{\mathbb{R}}\exp(-|x|)\,d\lambda\le4$: by monotonicity of the integral on each interval $[k,k+1)$ (where $\exp(-x)\le\exp(-k)$, a constant on an interval of Lebesgue measure one), the simple-function bounds and Monotone Convergence Theorem along $\mathbf{1}_{[0,L)}\exp(-x)\uparrow\mathbf{1}_{[0,\infty)}\exp(-x)$ give $\int_{[0,\infty)}\exp(-x)\,d\lambda\le\sum_{k\ge0}\exp(-1)^{k}\le2$ (geometric series; $\exp(1)\ge2$ from the series, so $\exp(-1)\le1/2$). The negative half-line contributes the same: the Preliminary with $\alpha=-1$, $\beta=0$, applied to $h(x)=\mathbf{1}_{[0,\infty)}(x)\exp(-x)$, for which $h(-x)=\mathbf{1}_{(-\infty,0]}(x)\exp(-|x|)$, gives $\int_{(-\infty,0]}\exp(-|x|)\,d\lambda=\int_{[0,\infty)}\exp(-x)\,d\lambda\le2$. Since $\exp(-|x|)\le\mathbf{1}_{(-\infty,0]}(x)\exp(-|x|)+\mathbf{1}_{[0,\infty)}(x)\exp(-x)$ pointwise (the point $0$ is counted in both halves, which only enlarges the bound), the two halves give $\int\exp(-|x|)\,d\lambda\le4$. Hence, splitting $\mathbb{R}$ into $[-2,2]$ and its complement and using $g\le1$,

$$c=\int g\,d\lambda\ \le\ \lambda([-2,2])+\int\exp(-|x|)\,d\lambda\ \le\ 4+4\ <\ \infty .$$

Also $c>0$: $g$ is nonincreasing in $|x|$, so $g\ge g(1)=\exp(-1/2)>0$ on $[-1,1]$, whence $c\ge\exp(-1/2)\,\lambda([-1,1])=2\exp(-1/2)>0$ by monotonicity. Finally $N=\nu/c$ is a measure (additivity is preserved under multiplication by the constant $1/c$) with $N(\mathbb{R})=c/c=1$, a probability measure.

Claim 3. For $s<t$, additivity gives $\nu((-\infty,t])-\nu((-\infty,s])=\nu((s,t])=\int\mathbf{1}_{(s,t]}\,g\,d\lambda\le\int\mathbf{1}_{(s,t]}\,d\lambda=t-s$, using $g\le1$, monotonicity, and interval lengths. So $t\mapsto\nu((-\infty,t])$ satisfies $|\nu((-\infty,t])-\nu((-\infty,s])|\le|t-s|$ and is continuous at every point (take $\delta=\varepsilon$); dividing by $c$, $\Phi$ is continuous on $\mathbb{R}$.

Claim 4. Write $g_2(x,y)=g(x)g(y)$, jointly Borel: coordinate projections are jointly Borel and compositions with Borel maps remain jointly Borel (claim 4 of Joint Distribution, Expectations, and Block Independence for Independent Random Variables), and products of jointly Borel functions are jointly Borel since $uv=((u+v)^{2}-u^{2}-v^{2})/2$.

(i) $c^{2}$ as a plane integral. $\lambda$ is $\sigma$-finite ($\mathbb{R}=\bigcup_L[-L,L]$), so Tonelli's and Fubini's Theorems applies to $g_2\ge0$: for fixed $y$, $\int g(x)g(y)\,d\lambda(x)=g(y)\,c$ (nonnegative scalars pull out, claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion), hence

$$\int_{\mathbb{R}^2}g_2\,d(\lambda\otimes\lambda)=\int_{\mathbb{R}}c\,g(y)\,d\lambda(y)=c^{2}.$$

(ii) Layer-cake identity. Let $(W,\mathcal{A},m)$ be a $\sigma$-finite measure space and $h:W\to[0,\infty]$ an $\mathcal{A}$-measurable function. Then

$$\int_Wh\,dm=\int_{(0,\infty)}m(\{h>s\})\,d\lambda(s).$$

Indeed, by density of the rationals the set $E_h=\{(w,s):0<s<h(w)\}$ satisfies $E_h=\bigcup_{q\in\mathbb{Q},q>0}\bigl(\{h>q\}\times(0,q)\bigr)$, a countable union of measurable rectangles, so $E_h\in\mathcal{A}\otimes\mathcal{B}(\mathbb{R})$ (Measurable Rectangles and the Product σ-Algebra). Applying Tonelli's and Fubini's Theorems to $\mathbf{1}_{E_h}$ over $m\otimes\lambda$: the $w$-indexed sections are $(E_h)_w=(0,h(w))$ with $\lambda((0,h(w)))=h(w)$ (interval lengths, including the value $\infty$ for $h(w)=\infty$, since $\lambda((0,\infty))=\infty$ by continuity from below), while the $s$-indexed sections are $\{h>s\}$ for $s>0$ and $\emptyset$ otherwise; the two iterated integrals give the two sides.

(iii) Dilations of the plane. For $r>0$ and $E\in\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$, write $rE=\{(rx,ry):(x,y)\in E\}$. The map $S_r(x,y)=(x/r,y/r)$ satisfies $S_r^{-1}(A\times B)=(rA)\times(rB)$, with $rA$ Borel (it is the preimage of $A$ under the continuous map $x\mapsto x/r$), so $S_r$ is measurable and $rE=S_r^{-1}(E)$ is again in the product $\sigma$-algebra. The set function $E\mapsto(\lambda\otimes\lambda)(rE)$ is a measure (preimages preserve disjoint unions) assigning to a rectangle the value $\lambda(rA)\lambda(rB)=r^{2}\lambda(A)\lambda(B)$, by the Preliminary with $\alpha=r$, $\beta=0$. The set function $E\mapsto r^{2}(\lambda\otimes\lambda)(E)$ is a measure with the same rectangle values, namely those of the product of the $\sigma$-finite measure $r\lambda$ with itself; by the uniqueness clause of Existence and Uniqueness of the Product Measure both equal $(r\lambda)\otimes(r\lambda)$, so

$$(\lambda\otimes\lambda)(rE)=r^{2}\,(\lambda\otimes\lambda)(E).$$

(iv) Disks. Let $D^{\circ}=\{(x,y):x^{2}+y^{2}<1\}$, which is in the product $\sigma$-algebra since $(x,y)\mapsto x^{2}+y^{2}$ is jointly Borel; similarly $D$ (use $\le$). For $0<s<1$, $sD\subseteq D^{\circ}\subseteq D$, so by monotonicity and (iii), $s^{2}\pi\le(\lambda\otimes\lambda)(D^{\circ})\le\pi$; letting $s\to1$ gives $(\lambda\otimes\lambda)(D^{\circ})=\pi$. Hence for $R>0$, $(\lambda\otimes\lambda)(R\,D^{\circ})=\pi R^{2}$.

(v) Conclusion. For $0<s<1$: $g_2(x,y)>s\iff\exp(-(x^{2}+y^{2})/2)>s\iff-(x^{2}+y^{2})/2>\ln s\iff x^{2}+y^{2}<2\ln(1/s)$, using the inverse relation between $\exp$ and $\ln$ and $-\ln s=\ln(1/s)$; thus $\{g_2>s\}=R_s\,D^{\circ}$ with $R_s=\sqrt{2\ln(1/s)}$ (positive square root), of measure $\pi R_s^{2}=2\pi\ln(1/s)$ by (iv). For $s\ge1$, $\{g_2>s\}=\emptyset$ since $g_2\le1$. By (i) and (ii) applied on $(\mathbb{R}^2,\lambda\otimes\lambda)$ ($\sigma$-finite by Existence and Uniqueness of the Product Measure),

$$c^{2}=\int_{(0,\infty)}(\lambda\otimes\lambda)(\{g_2>s\})\,d\lambda(s)=2\pi\int_{(0,1)}\ln(1/s)\,d\lambda(s).$$

Finally $\int_{(0,1)}\ln(1/s)\,d\lambda(s)=1$: by (ii) applied on $(\mathbb{R},\lambda)$ to $h(s)=\ln(1/s)\mathbf{1}_{(0,1)}(s)$ (measurable: for $u>0$, $\{h>u\}=\{s\in(0,1):1/s>\exp(u)\}=(0,\exp(-u))$, an interval; for $u\le 0$, $\{h>u\}\in\mathcal{B}(\mathbb{R})$ similarly),

$$\int_{(0,1)}\ln(1/s)\,d\lambda(s)=\int_{(0,\infty)}\lambda\bigl((0,\exp(-u))\bigr)\,d\lambda(u)=\int_{(0,\infty)}\exp(-u)\,d\lambda(u),$$

and the map $u\mapsto-\exp(-u)$ is an antiderivative of $u\mapsto\exp(-u)$ on every $[0,L]$ (one-dimensional chain rule with $\exp'=\exp$), so by Fundamental Theorem of Calculus, Part II in One Dimension and Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval, $\int\mathbf{1}_{[0,L]}\exp(-u)\,d\lambda=1-\exp(-L)$; by Monotone Convergence Theorem and $\exp(-L)\le1/L\to0$ (from $\exp(L)\ge L$), the integral equals $1$. Hence $c^{2}=2\pi$. $\blacksquare$