[proofplan]
The proof separates the sign randomness from the magnitude of $X$. Since a [Rademacher random variable](/page/Rademacher%20Random%20Variable) takes only the values $1$ and $-1$ almost surely, multiplication by $\varepsilon$ does not change absolute values outside a null set. For the distributional identity, we partition the event $\{\varepsilon X\in A\}$ according to the two sign events $\{\varepsilon=1\}$ and $\{\varepsilon=-1\}$, then use independence to factor the two probabilities. Applying the same decomposition to $-\varepsilon X$ gives the symmetry in distribution.
[/proofplan]
[step:Use the Rademacher values to preserve absolute values almost surely]
Define the event $N\in\mathcal F$ by
\begin{align*}
N=\{\omega\in\Omega:\varepsilon(\omega)\notin\{-1,1\}\}.
\end{align*}
Since $\varepsilon$ is Rademacher,
\begin{align*}
\mathbb P(N)=1-\mathbb P(\varepsilon=1)-\mathbb P(\varepsilon=-1)=0.
\end{align*}
For every $\omega\in\Omega\setminus N$, we have $|\varepsilon(\omega)|=1$, and hence
\begin{align*}
|\varepsilon(\omega)X(\omega)|=|\varepsilon(\omega)|\,|X(\omega)|=|X(\omega)|.
\end{align*}
Therefore $|\varepsilon X|=|X|$ almost surely.
[/step]
[step:Decompose the event $\{\varepsilon X\in A\}$ by the two sign events]
Fix a Borel set $A\in\mathcal B(\mathbb R)$. Define the events $E_+,E_-,C_A,D_A\in\mathcal F$ by
\begin{align*}
E_+=\{\omega\in\Omega:\varepsilon(\omega)=1\}
\end{align*}
\begin{align*}
E_-=\{\omega\in\Omega:\varepsilon(\omega)=-1\}
\end{align*}
\begin{align*}
C_A=\{\omega\in\Omega:X(\omega)\in A\}.
\end{align*}
Let $r:\mathbb R\to\mathbb R$ be the continuous map $r(x)=-x$, and define the Borel set $-A:=r^{-1}(A)$. Define
\begin{align*}
D_A=\{\omega\in\Omega:X(\omega)\in -A\}=\{\omega\in\Omega:-X(\omega)\in A\}.
\end{align*}
The events $E_+\cap C_A$ and $E_-\cap D_A$ are disjoint. For events $B,C\in\mathcal F$, write $B\triangle C:=(B\setminus C)\cup(C\setminus B)$ for their symmetric difference. Moreover,
\begin{align*}
\{\omega\in\Omega:\varepsilon(\omega)X(\omega)\in A\}\triangle\bigl((E_+\cap C_A)\cup(E_-\cap D_A)\bigr)\subset N.
\end{align*}
Since $\mathbb P(N)=0$, finite additivity over disjoint events gives
\begin{align*}
\mathbb P(\varepsilon X\in A)=\mathbb P(E_+\cap C_A)+\mathbb P(E_-\cap D_A).
\end{align*}
[guided]
We want to compute the probability that the signed [random variable](/page/Random%20Variable) $\varepsilon X$ lands in $A$. The sign $\varepsilon$ has only two possible values outside the null event
\begin{align*}
N=\{\omega\in\Omega:\varepsilon(\omega)\notin\{-1,1\}\}.
\end{align*}
Therefore the natural partition is by whether $\varepsilon=1$ or $\varepsilon=-1$.
For the fixed Borel set $A\in\mathcal B(\mathbb R)$, define
\begin{align*}
E_+=\{\omega\in\Omega:\varepsilon(\omega)=1\}
\end{align*}
and
\begin{align*}
E_-=\{\omega\in\Omega:\varepsilon(\omega)=-1\}.
\end{align*}
Also define
\begin{align*}
C_A=\{\omega\in\Omega:X(\omega)\in A\}.
\end{align*}
Let $r:\mathbb R\to\mathbb R$ be the continuous map $r(x)=-x$, and define $-A:=r^{-1}(A)$. Since $A$ is Borel and $r$ is continuous, $-A\in\mathcal B(\mathbb R)$. Define
\begin{align*}
D_A=\{\omega\in\Omega:X(\omega)\in -A\}=\{\omega\in\Omega:-X(\omega)\in A\}.
\end{align*}
These are events in $\mathcal F$ because $\varepsilon$ and $X$ are real-valued random variables, $A$ is Borel, and $-A$ is Borel.
On $E_+$, the condition $\varepsilon X\in A$ is exactly the condition $X\in A$, so the relevant part of the event is $E_+\cap C_A$. On $E_-$, the condition $\varepsilon X\in A$ is exactly the condition $-X\in A$, so the relevant part is $E_-\cap D_A$. Outside $E_+\cup E_-$ there may be no such simplification, but that exceptional set is precisely contained in $N$, which has probability zero. For events $B,C\in\mathcal F$, write $B\triangle C:=(B\setminus C)\cup(C\setminus B)$ for their symmetric difference. Thus
\begin{align*}
\{\omega\in\Omega:\varepsilon(\omega)X(\omega)\in A\}\triangle\bigl((E_+\cap C_A)\cup(E_-\cap D_A)\bigr)\subset N.
\end{align*}
The two events $E_+\cap C_A$ and $E_-\cap D_A$ are disjoint because $E_+\cap E_-=\varnothing$. Since changing an event by a null set does not change its probability, and since probability is finitely additive on disjoint unions, we obtain
\begin{align*}
\mathbb P(\varepsilon X\in A)=\mathbb P(E_+\cap C_A)+\mathbb P(E_-\cap D_A).
\end{align*}
[/guided]
[/step]
[step:Factor the two probabilities using independence]
Since $\{1\}\in\mathcal B(\mathbb R)$ and $A\in\mathcal B(\mathbb R)$, the stated independence of $\varepsilon$ and $X$ gives
\begin{align*}
\mathbb P(E_+\cap C_A)=\mathbb P(\varepsilon=1)\mathbb P(X\in A)=\frac{1}{2}\mathbb P(X\in A).
\end{align*}
Since $\{-1\}\in\mathcal B(\mathbb R)$ and $-A\in\mathcal B(\mathbb R)$, the same independence gives
\begin{align*}
\mathbb P(E_-\cap D_A)=\mathbb P(\varepsilon=-1)\mathbb P(X\in -A)=\frac{1}{2}\mathbb P(-X\in A).
\end{align*}
Substituting these two identities into the decomposition from the previous step yields
\begin{align*}
\mathbb P(\varepsilon X\in A)=\frac{1}{2}\mathbb P(X\in A)+\frac{1}{2}\mathbb P(-X\in A).
\end{align*}
[/step]
[step:Apply the same decomposition to prove symmetry in distribution]
Using the same fixed Borel set $A\in\mathcal B(\mathbb R)$, decompose the event $\{-\varepsilon X\in A\}$ over $E_+$ and $E_-$. On $E_+$ this event is $D_A$, and on $E_-$ this event is $C_A$, again up to the null set $N$. Hence
\begin{align*}
\mathbb P(-\varepsilon X\in A)=\mathbb P(E_+\cap D_A)+\mathbb P(E_-\cap C_A).
\end{align*}
By independence,
\begin{align*}
\mathbb P(E_+\cap D_A)=\frac{1}{2}\mathbb P(-X\in A)
\end{align*}
and
\begin{align*}
\mathbb P(E_-\cap C_A)=\frac{1}{2}\mathbb P(X\in A).
\end{align*}
Therefore
\begin{align*}
\mathbb P(-\varepsilon X\in A)=\frac{1}{2}\mathbb P(-X\in A)+\frac{1}{2}\mathbb P(X\in A)=\mathbb P(\varepsilon X\in A).
\end{align*}
Since this equality holds for every Borel set $A\in\mathcal B(\mathbb R)$, the real-valued random variables $\varepsilon X$ and $-\varepsilon X$ have the same distribution.
[/step]
