Isometry isomorphism between $L_1[0,1]$ and $L_1([0,1]^2)$ 
Question: Does there exist an isometry isomorphism between $L_1[0,1]$ and $L_1([0,1]^2)$?

I have a gut feeling that the answer is yes. But I do not how to show it.
Any hint is appreciated. 
 A: There exists  a map $f: [0,1] \to [0,1] \times [0,1]$ which is bijective such that $f$  and$f^{-1}$ are both Borel measurable and $m_1(E)=m_2(f(E))$ for every Borle set $E$ in $[0,1]$. Here $m_1$ and $m_2$ are the one and two dimensional Lebesgue measures respectively.
If $g \in L^{p}(m_1)$ define $G$ on $[0,1] \times [0,1]$ by $G(y)=g(f^{-1}(y))$. For  a simple function $g$ a routine verification shows that $\int |g|^{p} dm_1=\int |G|^{p} dm_2$. Hence the same equation holds for all Borel measurable functions $g$. The map $g \mapsto G$ is thus an isometric isomorphism from  $L^{p}(m_1)$ onto $L^{p}(m_2)$.
Reference for isomorphism theorem: Proposition 26.6 , page 118 of Introduction to Probability and Measure by K R Parthasarathy.
Note: Even a weaker form of the isomorphism theorem is good enough here. Instead of  a point isomorphism, a measure algebra isomorphism is good enough to show that the two $L^{p}$ spaces are isometrically isomorphic. 
In the following  post on MSE you will see another reference to the isomorphism Theorem  : There is only one interesting measure space
A: Consider (inverse of) the Cantor map $\phi\colon [0,1]\times [0,1]\to [0,1]$ given in dyadic expansion
$$(0.a_1 a_2 \ldots, 0.b_1 b_2\ldots)\mapsto 0.a_1 b_1 a_2 b_2 \ldots$$
In fact there exist $Z\subset [0,1]\times[0,1]$, $Z'\subset [0,1]$ measure $0$ sets so that the map $\phi \colon [0,1]^2 \backslash Z\to [0,1]\backslash Z'$ is a homeomorphism that preserves the measure. 
The map (classes from Borel measurable on $[0,1]$) to (classes of Borel measurable functions on $[0,1]\times [0,1]$) takes a function $f$ to :
$$\phi^{\star}(f) (0.a_1 a_2 \ldots, 0.b_1 b_2\ldots)=f(0.a_1 b_1 a_2 b_2 \ldots)$$
This will give the required isometry.
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