$L^2$ Space of the Unit Circle and Bilateral Shift I'm trying to understand the space $L^2(\Bbb{T},m)$ or $L^2(\Bbb{T},\frac{d \theta}{2 \pi})$, where $\Bbb{T}$ is the unit circle and $m$ is the normalized Lebesgue measure; and the I'm trying to understand the Bilateral shift on this space. Specifically, I am trying to understand them as they are presented in the papers paper#1 and paper#2 
From my understanding, 
$$L^2 (\Bbb{T},m) = \{f : \Bbb{T} \to \Bbb{C} \mid f \text{ is measurable and } \int_{\Bbb{T}} |f|^2 dm < \infty \},$$
and $L^2(\Bbb{T},\frac{d \theta}{2 \pi})$ equals the same thing but with $dm$ replaced by $\frac{d \theta}{2 \pi}$. Yet in paper#1 it is claimed that $\{w^n \mid n \in \Bbb{Z}, |w| = 1 \}$ is an orthonormal basis for $L^2(\Bbb{T},m)$ (see example 3). But how can that be? The set just consists of complex numbers, not functions. 
Also, it seems like $\Bbb{T}$ is being identified with $[0,2 \pi]$, where $0$ and $2 \pi$ are being glued together; and integration is being defined in terms of this identification...but I'm not entirely sure...This doesn't synch up with paper#2, where the bilateral shift is defined (see page 8). It is that operator which multiplies functions on the left by $e^{i \theta}$ for some $\theta$. I'm having trouble understanding how $\Bbb{T}$, $L^2(\Bbb{T},m)$ (and $L^2(\Bbb{T}, \frac{d \theta}{2 \pi})$, and the Bilateral shift are defined; and how all of these perspectives/pictures synch up. 
It seems like the authors are being sloppy with notation and not entirely rigorous. Is there anyway to put these ideas on a firmer foundation?
 A: When they write 
$$\tag1
\{w^n:\ n\in\mathbb Z,\ |w|=1\}
$$
and say it is an orthonormal basis, that's clearly a typo. If you consider the Lebesgue measure on the unit circle, normalized (that is, divided by $2\pi$ so that the measure of the circle is $1$), then the canonical orthonormal basis is $\{w^n\}_n$, where $w:\mathbb T\to\mathbb T$ is the identity function $w(z)=z$. So we can write the canonical orthonormal basis as 
$$\tag2
\{w^n:\ n\in\mathbb Z\}.
$$
If I had to speculate, the typo comes from writing $\mathbb T$ as in $(1)$ (note that it would be coherent if you change the period for an equal sign), and this got mixed into the next sentence, where something like $(2)$ was wanted. 
In general, if you fix an orthonormal basis $\{e_n\}_{n\in\mathbb Z}$, the bilateral shift corresponding to that basis is the linear map $B$ induced by $e_n\to e_{n+1}$. Now, with the particular basis in $(2)$, the bilateral shift maps $w^n$ to $w^{n+1}$. If you evaluate at an arbitrary point $z=e^{i\theta}$, then 
$$
(Bw^n)(z)=(w^{n+1})(z)=e^{i(n+1)\theta}=e^{i\theta}\,e^{in\theta}=z\,w^n(z).
$$
So the bilateral shift consists of multiplication by the identity function: the above computation is made on the elements of the basis, but it extends by linearity and continuity to all of $L^2(\mathbb T)$: so $(Bf)(z)=zf(z)$. When they say "multiplication by $e^{i\theta}$, then don't mean multiplication by a fixed complex number, but by the identity function. 
