Density normal laws and square-summable Gaussian series under countable choice

The positive continuous function $g$ is Borel measurable, since inverse images of open sets are open. For an integer $R\ge 1$, substitute $x=\sqrt{2}t$ in the continuous compact-interval integral. The bridge in [F1] and then increasing truncation in [F2] give ${\int }_{\mathbb{R}}{e}^{-x^2\mathrm{/}2}dx=\sqrt{2}{\lim }_{R\to \mathrm{\infty }}{\int }_{-R\mathrm{/}\sqrt{2}}^{R\mathrm{/}\sqrt{2}}{e}^{-t^2}dt=\sqrt{2\pi }$. Thus [F2] makes $\gamma$ a probability. For $v>0$, compact substitutions in the same way show that its affine pushforward has the asserted half-line integrals of $g_{m,v}$; limits of compact intervals give the half-lines, and [F5] gives equality of the measures. For $v=0$ the affine map is constant, giving the Dirac law directly.

The CDF of $N(m,v)$ for $v>0$ is continuous, since $g_{m,v}$ is bounded and the integral over an interval of length $h$ is at most $\mathrm{\mid }h\mathrm{\mid }\mathrm{/}\sqrt{2\pi v}$. It is strictly increasing because the positive continuous density has a positive minimum on every nondegenerate compact interval. For $Z$ with density $g$, $g^{\mathrm{\prime }}(x)=-xg(x)$, and integration by parts gives ${\int }_{-R}^R\mathrm{\mid }x\mathrm{\mid }g(x)dx=2(g(0)-g(R))$ and ${\int }_{-R}^R{x}^{2}g(x)dx={\int }_{-R}^{R}g(x)dx-2Rg(R)$. The exponential series shows $g(R)\to 0$ and $Rg(R)\to 0$. Increasing truncation gives $\mathbb{E}\mathrm{\mid }Z\mathrm{\mid }<\mathrm{\infty }$ and $\mathbb{E}{Z}^2=1$; symmetry gives $\mathbb{E}Z=0$. Expanding $m+\sqrt{v}Z$ gives the claimed mean and variance, also for $v=0$.

Let $U,V$ be independent centered density normal variables with variances $a,b>0$. Their sum has density $h(x)={\int }_{\mathbb{R}}{g}_{0,a}(u)g_{0,b}(x-u)du$. Indeed, [F4] expresses $P(U+V\le x)$ as the nonnegative double integral of ${\mathbf{1}}_{u+t\le x}{g}_{0,a}(u)g_{0,b}(t)$; for each fixed $u$, a translation in the continuous half-line integral followed by Tonelli gives ${\int }_{-\mathrm{\infty }}^{x}h(y)dy$. These half-line probabilities determine the law by [F5]. Completing the square gives $\frac{u^2}{a}+\frac{(x-u{)}^2}{b}=\frac{a+b}{ab}{(u-\frac{ax}{a+b})}^2+\frac{x^2}{a+b}\mathrm{.}$ The compact substitutions and increasing limits in step 1.1 therefore give $h(x)=g_{0,a+b}(x)$. If either variance is zero the corresponding variable is zero almost surely and the same conclusion is immediate. Scaling a coordinate by a negative coefficient gives the same density as scaling by its absolute value, by the symmetry of $g$. Finite induction, using mutual independence to separate the last coordinate from the preceding tuple, proves $S_m\sim N(0,v_m)$ with $v_m={\sum }_{j\le m}{c}_j^2$.

If $v=0$, every coefficient is zero, so convergence in probability implies $S=0$ almost surely. Suppose $v>0$. Then $v_m\to v$ and eventually $v_m>0$. Write $\mathrm{\Phi }$ for the continuous CDF of $\gamma$. For any $x\in \mathbb{R}$ and $\epsilon >0$, convergence in probability and step 3.1 give $\mathrm{\Phi }((x-\epsilon )\mathrm{/}\sqrt{v_m})-P(\mathrm{\mid }{S}_m-S\mathrm{\mid }>\epsilon )\le P(S\le x)\le \mathrm{\Phi }((x+\epsilon )\mathrm{/}\sqrt{v_m})+P(\mathrm{\mid }{S}_m-S\mathrm{\mid }>\epsilon )\mathrm{.}$ First let $m\to \mathrm{\infty }$, then $\epsilon \downarrow 0$. Continuity from step 2.1 gives $P(S\le x)=\mathrm{\Phi }(x\mathrm{/}\sqrt{v})$ for every $x$. By [F5], $S$ has law $N(0,v)$. Almost sure convergence implies convergence in probability by dominated convergence on the indicators of $\{\mathrm{\mid }{S}_m-S\mathrm{\mid }>\epsilon \}$, and $L^2$ convergence implies it by ${\epsilon }^{2}P(\mathrm{\mid }{S}_m-S\mathrm{\mid }>\epsilon )\le \mathbb{E}\mathrm{\mid }{S}_m-S{\mathrm{\mid }}^2$.