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The lecture notes say:

Let $(\Omega,\mathcal{A},P) = ((0,1],\mathcal{B}((0,1]),\lambda)$ where $\lambda$ is the Lebesgue measure on the unit interval.

Define $X(\omega) = 1$ for $\omega > 1/2$ and $X(\omega) = - 1$ for $\omega \le 1/2$. Then $T(\omega) = 2\omega - [2\omega]$.

It can be shown that $X_{k} = \pm 1$ with prob. $1/2$ is an iid Rademacher sequence.

Can someone please outline how this is possible?

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    $\begingroup$ Your description is sufficient enough to generate them. Start with generating a random variable $\omega$ from $\mathcal{U}((0, 1])$. $\endgroup$ – Zhanxiong Jun 1 '18 at 13:23
  • $\begingroup$ @Zhanxiong that begs the question. Has $\omega$ been defined? $\endgroup$ – AdamO Jun 1 '18 at 15:35
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This is just a change of variable calculation. Use the CDF method. The CDF of a Rademacher RV is just $0$ if $x < -1$, $0.5$ if $-1 < x < 1$ and $1$ if $x > 1$. You can show $2\omega - [2\omega]$ satisfies this when $\omega \sim \text{Uniform}(0,1)$.

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