What is the Lebesgue measure on $\mathbb{T}$? This is a rather basic question in complex analysis. Let $\mathbb{T}$  be the complex unit circle. Now, I am trying to understand what is the Lebesgue probability measure on $\mathbb{T}$. To my understanding, we take the Lebesgue measure on the segment $[0,2\pi]$ and we identify $\mathbb{T}$ with $[0,2\pi]$ by the map $t\mapsto e^{it}$. So if $\eta$ is Lebesgue measure on $\mathbb{T}$, we have $\int_\mathbb{T}f(z)d\eta(z)=\frac1{2\pi}\int_0^{2\pi}f(e^{it})dt$
. Did I get it right?
 A: I am not sure what is meant by the Lebesgue measure on the $\Bbb S$ as I've seen such concept defined only for $\Bbb R^n$, but you can certainly do what you did. More generally, starting with the Lebesgue measure $\lambda$, you can further normalize it and define a probability measure $\nu = \frac{1}{2\pi}\lambda$ on $[0,2\pi]$. By choosing any measurable map $g:[0,2\pi]\to \Bbb S$ which is not necessarily bijective, you obtain an image measure
$
  \mu_g:=\nu\circ g^{-1}
$
on the space $\Bbb S$. 
By the definition of the image measure, it holds that
$$
  \mu_g(A) = \nu(\{t\in [0,2\pi]:g(t)\in A\}) = \nu(g^{-1}(A))
$$
for any measurable subset $A$ of $\Bbb S$ and as a result
$$
  \int_\Bbb S f(s)\mu_g(\mathrm ds) = \int_0^{2\pi}f(g(t)) \nu(\mathrm dt) = \frac{1}{2\pi}\int_0^{2\pi}f(g(t)) \mathrm dt
$$
for any bounded measurable $f$, which you easily get by approximating $f$ with simple functions.
On the other hand, if you choose the map $g(t) = \mathrm e^{it}$, the correspondent measure $\mu_g$ on the sphere has certain nice features. In particular, it is shift invariant.
