[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).[/step]

[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: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]