Independent identical distribution sequence given a measurable function $\alpha: (\Omega, \mu) \to \mathbb{R}$, and transformation $\sigma : \Omega \to \Omega $. 
I found an example such that $\alpha, \alpha\circ \sigma, \alpha\circ \sigma^2, \alpha\circ \sigma^3 \cdots : \Omega \to \Omega$ are all Independent identical distribution :
Let $(\Omega, \mu) = ([0,1], Leb)$, $\alpha $ takes countable $0-1$ value, and $\sigma$ is piecewise linear map and some $r \in (0,1)$ such that:
$$ \alpha(\omega)=\left\{
\begin{aligned}
0 &  & \omega \in [0,r] \\
1 &  & \omega \in [r, 1]\\
\end{aligned}
\right.
$$
$$ \sigma(\omega)=\left\{
\begin{aligned}
\frac{\omega}{r} &  & \omega \in [0,r] \\
\frac{\omega-r}{1-r}&  & \omega \in [r, 1]\\
\end{aligned}
\right.
$$
We can also construct similar example such that $\alpha$ takes countable values, i.e. distribution of $\alpha$ has only countable atoms.
Can we construct an example on $\Omega= [0,1] $such that 
1, $\alpha, \alpha\circ \sigma, \alpha\circ \sigma^2, \alpha\circ \sigma^3 \cdots $ are Independent identical distribution and
2, distribution of $\alpha$ has no atom and absolutely continuous w.r.t Leb? Thanks in advanced!
 A: The easiest way to do this is to take $\Omega=\{[0,1)\}^{\mathbb N_0}$., $\sigma$ the shift map and $\alpha(\omega)=\omega_0$.
If you don’t like the infinite-dimensional $\Omega$, you can build an example with $\Omega=[0,1)$ at the cost of making the transformation uglier (there is an measure space isomorphism mapping $[0,1)^{\mathbb N_0}$ equipped with the product of Lebesgue measures to $[0,1)$ equipped with Lebesgue, and if you conjugate $\sigma$ by this measure space isomorphism, you obtain a transformation on $[0,1)$ with the desired property. 
Let me add: in some sense you should not expect a better answer than this. You seem to have edited the question to require $\Omega=[0,1]$. In this case, the map $\Phi$ from $\Omega$ to $\mathbb R^{\mathbb N_0}$ defined by $\omega\mapsto (\alpha(\sigma^n\omega))_{n\in\mathbb N_0}$ is a factor map. Since its range has infinite entropy, it follows that $\sigma\colon\Omega\to\Omega$ has infinte entropy. This can never be obtained by piecewise smooth maps $\sigma$ like your example.
