Isometry isomorphism between $L_1[0,1]$ and $L_1([0,1]^2)$

Question

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.

Existence follows by isomorphism theorem: the spaces $[0,1]$ and $[0,1] \times [0,1]$ with Lebesgue measure are isomorphic so the corresponding $L^{p}$ spaces are isometrically isomorphic for every $p$. I do not know if we can write down an explicit isometrically isomorphism. — Kavi Rama Murthy, Feb 25 '20 at 05:34
@KaviRamaMurthy May I know what kind of isomorphism between $[0,1]$ and $[0,1]\times [0,1]?$ Isometric isomorphism? — Idonknow, Feb 25 '20 at 05:35
Measure Theory by Cohn has a detailed discussion of isomorphisms between measure spaces. In the present case there is a bijective map which preserves measurability and measures of sets. — Kavi Rama Murthy, Feb 25 '20 at 05:37
@KaviRamaMurthy Thanks for the reference. If possible, can you point me to the specific chapter/section? — Idonknow, Feb 25 '20 at 05:40
I guess so. My copy was published in 1980 and no edition number is mentioned. The theorem is in the chapter on Polish spaces and Analytic sets. — Kavi Rama Murthy, Feb 25 '20 at 06:31
@Idonknow why is there a bounty on this? didn't Kavi answer? — mathworker21, Feb 27 '20 at 05:04
@mathworker21 it would be good if someone can explicitly mention how to use Theorem $8.3.6$, p. $275$ to solve my posted question as I do not see how. — Idonknow, Feb 27 '20 at 05:07

Kavi Rama Murthy · Accepted Answer · 2020-02-27T06:19:42.317

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

I feel like your note is wrong, even if I don't have a rational reason to think so. — mathworker21, Feb 27 '20 at 05:41
@mathworker21 Map $\sum a_i I_{E_i}$ to $\sum a_i I_{f (E_i)}$, show that this is an isometry and extend it to the whole of $L^{p}$. I am pretty sure this is correct. — Kavi Rama Murthy, Feb 27 '20 at 05:44

score 1 · Answer 2 · answered Feb 27 '20 at 06:21

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. '

Isometry isomorphism between $L_1[0,1]$ and $L_1([0,1]^2)$

2 Answers2