What is the normalised arc length measure on the unit circle?

Question

Consider the following fragment from Murphy's '$C^*$-algebras and operator theory':

Can someone explain what the normalised arc length measure on $\Bbb{T}$ is? Is this some translation of Lebesgue measure on the circle?

Answer 1

Are you familiar with Haar measure?

If not, take the map $[0,1)\mapsto \mathbb{T}$ given by $t\mapsto e^{2\pi it}$. This is a bijection, and you can transfer the Lebesgue measure from $[0,1)$ onto the unit circle. Note that translation invariance on the unit interval translates (no pun intended) into invariance under multiplication on the unit circle.

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Haar measure is a generalization of Lebesgue measure to other (sufficiently well-behaved) groups. The structure $(\mathbb{R},+)$ is a locally compact Hausdorff group, and the Lebesgue measure on $\mathbb{R}$ is translation invariant, i.e. $$\lambda(x+A)=\lambda(A)$$ for any real $x$ and any measurable set $A$. The construction I outlined above yields a measure $m$ on the unit circle that satisfies $$ m(z·A)=m(A) $$ for every $z$ in the unit circle and every measurable set $A$. Note that $\mathbb{T}$ is also a locally compact Haussdorff group. Also, the measure $m$ is regular since the Lebesgue measure is. These properties characterize general Haar measures, see here for more details https://en.wikipedia.org/wiki/Haar_measure

If you know about Fourier analysis, looking at Fourier transforms on $\mathbb{T}$ and Fourier series of periodic functions can be unified under the same framework: you're studying the generalized Fourier transform with respect to different Haar measures. See Rudin's Fourier Analysis on Groups for more information.