The Gaussian Integral Is Finite and Positive, and the First Two Gaussian Moments

In the settings of Measurable Functions and the Lebesgue Integral: Notation and The Real Line: Standing Notation and Background for Calculus on Intervals, let $\lambda$ be Lebesgue measure on $\mathbb{R}$ and $\exp$ the exponential function, and let $g_{0},g_{1},g_{2}:\mathbb{R}\to\mathbb{R}$ be the functions

$$g_{0}(x)=\exp(-x^{2}/2),\qquad g_{1}(x)=x\exp(-x^{2}/2),\qquad g_{2}(x)=x^{2}\exp(-x^{2}/2).$$

Let $c_{\mathrm{G}}\in[0,\infty]$ be the integral of $g_{0}\in\mathcal{M}^{+}(\mathbb{R},\mathcal{B}(\mathbb{R}))$ ($g_{0}$ is continuous and positive, Basic Properties of the Exponential Function (viii), Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (iv), Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii)):

$$c_{\mathrm{G}}=\int_{\mathbb{R}}\exp(-x^{2}/2)\,d\lambda(x).$$

(i) The Gaussian integral is finite and positive. $g_{0}\in\mathcal{L}^{1}(\lambda)$ and $0<c_{\mathrm{G}}<\infty$.

(ii) First and second moments. $g_{1},g_{2}\in\mathcal{L}^{1}(\lambda)$, and

$$\int_{\mathbb{R}}x\exp(-x^{2}/2)\,d\lambda(x)=0,\qquad\int_{\mathbb{R}}x^{2}\exp(-x^{2}/2)\,d\lambda(x)=c_{\mathrm{G}}.$$