Difference between an $L^p$ space on a bounded set and a periodic $L^p$ space? I'm confused about the concept periodic $L^p$ space. Let $\mathbb{T}$ be the quotient space $\mathbb{R}/\mathbb{Z}$. Here are my questions:


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*What is the Lebesgue measure on $\mathbb{T}$? How do people assign measures on a quotient space? 

*What exactly is the difference between $L^p(\mathbb{T})$ and $L^p([0,1])$? 



What I think is that $L^p([0,1])$ is the restriction of $L^p(\mathbb{R})$ functions on $[0,1]$; while $L^p(\mathbb{T})$ is the space of periodic functions (with period $1$) $f$ on $\mathbb{R}$ such that $f1_{[0,1]}\in L^p([0,1])$. 
Could anyone come up with references for detailed explanation? 
 A: To quote Tao's comment on this question:

One can define the Lebesgue measure on ${\bf T}$ by identifying that space with one of its fundamental domains, such as $[0,1)$. The spaces ${\bf T}$ and $[0,1)$ are thus isomorphic as measure spaces, and so $L^p({\bf T})$ and $L^p([0,1))$ are isomorphic as normed vector spaces. However, ${\bf T}$ and $[0,1)$ differ in other respects if one considers other structures than the measurable structure. For instance, from the point of view of topological structure ${\bf T}$ is compact and not simply connected, while $[0,1)$ is non-compact and simply connected. The differential structure is also different, for instance the identity function $x \mapsto x$ on $[0,1)$ is smooth using the differential structure of $[0,1)$, but is not even continuous if one replaces the domain with ${\bf T}$. In particular, PDE on $[0,1)$ and on ${\bf T}$ are quite different.
If one only cares about the measurable structure, then $[0,1)$, $[0,1]$, and ${\bf T}$ are isomorphic (up to null sets), and if one only cares about the normed vector space structure, $L^p([0,1))$, $L^p([0,1])$, and $L^p({\bf T})$ are all isomorphic. However, for applications to PDE one usually needs more structure than just the measurable or normed vector space structure (in particular, one needs differential structure), and then the three domains are all inequivalent (for instance the Sobolev spaces on the three domains behave differently).

A: One possible answer to your question is the following:
The space $[0,1)$ forms, with respect to the subspace topology, a topological group under ordinary addition, and the space $\mathbb{T} := \mathbb{R} / \mathbb{Z}$ forms, with respect to the quotient topology, a topological group under addition modulo $1$. The map 
$$ \varphi : [0,1) \to \mathbb{T}, \quad x \mapsto e^{2\pi i x} $$
is an isomorphic homeomorphism between the topological groups $([0,1), +)$ and $(\mathbb{T}, +_1)$, and hence $\mathbb{T} \cong [0,1)$ as topological groups. From this it follows that $L^p ([0,1)) \cong L^p (\mathbb{T})$.
A slightly different approach regarding the identification of $\mathbb{T}$ and $[0,1)$ is given in $\S 3.1.1$, called the The n-Torus $\mathbb{T}^n$, of Grafakos' Classical Fourier Analysis.
