Proof of Reflection Invariance of Lebesgue Measure and Symmetry of the Standard Normal Distribution

Write $r:\mathbb{R}\to\mathbb{R}$, $r(x)=-x$, so that $r\circ r$ is the identity and $-A=r^{-1}(A)$ for every $A\subseteq\mathbb{R}$.

Claim 1. Let $A\subseteq\mathbb{R}$ and let $\bigl((a_m,b_m)\bigr)_{m\in\mathbb{N}}$ be a sequence of open intervals covering $A$ as in Lebesgue Outer Measure on the Real Line. Then the intervals $(-b_m,-a_m)$ cover $-A$: if $x\in-A$ then $-x\in A$, so $a_m<-x<b_m$ for some $m$, i.e., $-b_m<x<-a_m$; and each has the same length $(-a_m)-(-b_m)=b_m-a_m$. Thus every covering of $A$ induces a covering of $-A$ with the same total length, so $\lambda^{*}(-A)\le\lambda^{*}(A)$; applying this inequality to $-A$ in place of $A$ and using $-(-A)=A$ gives the reverse inequality, hence $\lambda^{*}(-A)=\lambda^{*}(A)$.

Next, the family $\mathcal{D}=\{B\subseteq\mathbb{R}:\ -B\in\mathcal{B}(\mathbb{R})\}$ is a $\sigma$-algebra: $-\mathbb{R}=\mathbb{R}$, $-(\mathbb{R}\setminus B)=\mathbb{R}\setminus(-B)$, and $-\bigcup_m B_m=\bigcup_m(-B_m)$. It contains every open subset $U$ of $\mathbb{R}$: if $x\in-U$ then $-x\in U$, so there is $\varepsilon>0$ with $(-x-\varepsilon,-x+\varepsilon)\subseteq U$, and then $(x-\varepsilon,x+\varepsilon)=-(-x-\varepsilon,-x+\varepsilon)\subseteq-U$, so $-U$ is open, hence Borel. Since $\mathcal{B}(\mathbb{R})$ is the $\sigma$-algebra generated by the open sets, $\mathcal{B}(\mathbb{R})\subseteq\mathcal{D}$; that is, $-B$ is Borel whenever $B$ is. Since $\lambda$ is the restriction of $\lambda^{*}$ to $\mathcal{B}(\mathbb{R})$ by Existence of Lebesgue Measure on the Real Line, $\lambda(-B)=\lambda^{*}(-B)=\lambda^{*}(B)=\lambda(B)$.

Claim 2. Let $g(x)=\exp(-x^{2}/2)$ as in Standard Normal Distribution, and note $g(-x)=g(x)$ for every $x$, since $(-x)^{2}=x^{2}$. We first show that for every measurable $f:\mathbb{R}\to\mathbb{R}$ (with respect to $\mathcal{B}(\mathbb{R})$) with $f\ge0$,

$$\int_{\mathbb{R}}f\circ r\,d\lambda=\int_{\mathbb{R}}f\,d\lambda .$$

First, $f\circ r$ is measurable: $(f\circ r)^{-1}(B)=-f^{-1}(B)$ is Borel by Claim 1. For a nonnegative simple function $s=\sum_{i}c_i\mathbf{1}_{A_i}$ with Borel $A_i$ and $c_i\ge0$, the composition $s\circ r=\sum_i c_i\mathbf{1}_{-A_i}$ is again a nonnegative simple function, and its integral is $\sum_i c_i\,\lambda(-A_i)=\sum_i c_i\,\lambda(A_i)$ by Claim 1; that is, simple functions and their reflections have equal integrals. Moreover $s\mapsto s\circ r$ is a bijection from the set of simple functions $s$ with $0\le s\le f\circ r$ pointwise onto the set of simple functions $s'$ with $0\le s'\le f$ pointwise: it is its own inverse because $r\circ r$ is the identity, and $0\le s\le f\circ r$ holds pointwise if and only if $0\le s\circ r\le f$ does. Hence the two suprema defining the integrals of $f\circ r$ and of $f$ range over equal sets of real numbers, and the displayed identity follows.

Now fix a Borel set $B$ and apply the identity to $f=\mathbf{1}_{B}\,g$, which is measurable (a product of measurable functions, by the preliminaries of Square-Integrable Random Variables and the Mean-Square Inner Product applied on the measurable space $(\mathbb{R},\mathcal{B}(\mathbb{R}))$) and nonnegative. Since $\mathbf{1}_{B}(-x)=\mathbf{1}_{-B}(x)$ and $g(-x)=g(x)$, we get $f\circ r=\mathbf{1}_{-B}\,g$, so

$$\int_{\mathbb{R}}\mathbf{1}_{-B}\,g\,d\lambda=\int_{\mathbb{R}}\mathbf{1}_{B}\,g\,d\lambda,$$

and dividing by the normalizing constant $c$ of Standard Normal Distribution gives $N(-B)=N(B)$.

Claim 3. The function $-Z=(-1)Z$ is a random variable by the closure preliminaries of Square-Integrable Random Variables and the Mean-Square Inner Product. For every Borel set $B$, the events $\{-Z\in B\}$ and $\{Z\in-B\}$ are equal, since $-Z(\omega)\in B$ holds exactly when $Z(\omega)\in-B$. Because the distribution of $Z$ is $N$ and $-B$ is Borel by Claim 1,

$$P(-Z\in B)=P(Z\in-B)=N(-B)=N(B)$$

by Claim 2. Hence the distribution of $-Z$ is $N$, i.e., $-Z$ is standard normal by Standard Normal Distribution. $\blacksquare$