[proofplan]
The proof is just the distributional effect of two affine changes of variables. First, the map $x\mapsto 2x-1$ sends the Bernoulli support points $0$ and $1$ to the Rademacher support points $-1$ and $1$, and the corresponding events have the same probabilities. Conversely, the map $x\mapsto (x+1)/2$ sends $-1$ and $1$ back to $0$ and $1$, so the point masses match the Bernoulli law with parameter $\frac{1}{2}$.
[/proofplan]
[step:Push the Bernoulli variable forward by $x\mapsto 2x-1$]
Let $T:\mathbb R\to\mathbb R$ be the Borel measurable affine map defined by $T(x)=2x-1$ for every $x\in\mathbb R$. Since $B$ is a real-valued [random variable](/page/Random%20Variable), $\varepsilon=T\circ B$ is a real-valued random variable.
The event on which $\varepsilon$ equals $1$ is
\begin{align*}
\{\omega\in\Omega:\varepsilon(\omega)=1\}=\{\omega\in\Omega:B(\omega)=1\}.
\end{align*}
Indeed, $2B(\omega)-1=1$ is equivalent to $B(\omega)=1$. Therefore
\begin{align*}
\mathbb P(\varepsilon=1)=\mathbb P(B=1)=\frac{1}{2}.
\end{align*}
Similarly, the event on which $\varepsilon$ equals $-1$ is
\begin{align*}
\{\omega\in\Omega:\varepsilon(\omega)=-1\}=\{\omega\in\Omega:B(\omega)=0\},
\end{align*}
because $2B(\omega)-1=-1$ is equivalent to $B(\omega)=0$. Hence
\begin{align*}
\mathbb P(\varepsilon=-1)=\mathbb P(B=0)=\frac{1}{2}.
\end{align*}
Since these two probabilities sum to $1$, we also have
\begin{align*}
\mathbb P(\varepsilon\in\{-1,1\})=1.
\end{align*}
Thus $\varepsilon$ has the symmetric distribution on $\{-1,1\}$, so $\varepsilon$ is a [Rademacher random variable](/page/Rademacher%20Random%20Variable).
[guided]
Let $T:\mathbb R\to\mathbb R$ be the affine map defined by $T(x)=2x-1$ for every $x\in\mathbb R$. This map is continuous, hence Borel measurable. Since $B:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable, the composition $\varepsilon=T\circ B$ is also measurable, so $\varepsilon$ is a real-valued random variable.
We now compute its point masses. The equation $\varepsilon(\omega)=1$ means
\begin{align*}
2B(\omega)-1=1.
\end{align*}
Solving this equation gives $B(\omega)=1$. Therefore the two events are equal:
\begin{align*}
\{\omega\in\Omega:\varepsilon(\omega)=1\}=\{\omega\in\Omega:B(\omega)=1\}.
\end{align*}
Taking probabilities and using the hypothesis on $B$, we get
\begin{align*}
\mathbb P(\varepsilon=1)=\mathbb P(B=1)=\frac{1}{2}.
\end{align*}
The same computation at the other support point gives the second mass. The equation $\varepsilon(\omega)=-1$ means
\begin{align*}
2B(\omega)-1=-1.
\end{align*}
Solving gives $B(\omega)=0$, so
\begin{align*}
\{\omega\in\Omega:\varepsilon(\omega)=-1\}=\{\omega\in\Omega:B(\omega)=0\}.
\end{align*}
Thus
\begin{align*}
\mathbb P(\varepsilon=-1)=\mathbb P(B=0)=\frac{1}{2}.
\end{align*}
Finally, the two events $\{\varepsilon=1\}$ and $\{\varepsilon=-1\}$ are disjoint, and their probabilities add to $1$. Hence
\begin{align*}
\mathbb P(\varepsilon\in\{-1,1\})=\mathbb P(\varepsilon=1)+\mathbb P(\varepsilon=-1)=1.
\end{align*}
So $\varepsilon$ is supported on $\{-1,1\}$ almost surely and gives each point probability $\frac{1}{2}$. This is exactly the Rademacher distribution.
[/guided]
[/step]
[step:Push the Rademacher variable back by $x\mapsto (x+1)/2$]
Let $U:\mathbb R\to\mathbb R$ be the Borel measurable affine map defined by $U(x)=(x+1)/2$ for every $x\in\mathbb R$. Since $\varepsilon$ is a real-valued random variable, $B=U\circ\varepsilon$ is a real-valued random variable.
Because $\varepsilon$ is Rademacher,
\begin{align*}
\mathbb P(\varepsilon=1)=\frac{1}{2}
\end{align*}
and
\begin{align*}
\mathbb P(\varepsilon=-1)=\frac{1}{2}.
\end{align*}
The event on which $B$ equals $1$ is
\begin{align*}
\{\omega\in\Omega:B(\omega)=1\}=\{\omega\in\Omega:\varepsilon(\omega)=1\},
\end{align*}
because $(\varepsilon(\omega)+1)/2=1$ is equivalent to $\varepsilon(\omega)=1$. Therefore
\begin{align*}
\mathbb P(B=1)=\frac{1}{2}.
\end{align*}
Likewise,
\begin{align*}
\{\omega\in\Omega:B(\omega)=0\}=\{\omega\in\Omega:\varepsilon(\omega)=-1\},
\end{align*}
so
\begin{align*}
\mathbb P(B=0)=\frac{1}{2}.
\end{align*}
Since these two probabilities sum to $1$, $B$ is supported on $\{0,1\}$ almost surely. Thus $B$ has Bernoulli distribution with parameter $p=\frac{1}{2}$.
[/step]
[step:Identify the two constructions as inverse conversions]
The affine maps $T(x)=2x-1$ and $U(x)=(x+1)/2$ exchange the two-point sets $\{0,1\}$ and $\{-1,1\}$. The preceding steps show that they preserve the corresponding point masses in the required way:
\begin{align*}
0\mapsto -1,\qquad 1\mapsto 1
\end{align*}
for the Bernoulli-to-Rademacher direction, and
\begin{align*}
-1\mapsto 0,\qquad 1\mapsto 1
\end{align*}
for the Rademacher-to-Bernoulli direction. Hence the first construction produces a Rademacher random variable, and the second construction produces a Bernoulli random variable with parameter $p=\frac{1}{2}$.
[/step]
