TITLE: Measurable homomorphism of $\mathbb{T}$ into $\mathbb{C}^\times$
QUESTION [6 upvotes]: Working through Katznelson's An Introduction to Harmonic Analysis and have been stumped by the following problem for the past few days: Show that a measurable homomorphism of $\mathbb{T}=\mathbb{R}/2\pi\mathbb{Z}$ into $\mathbb{C}^\times$ is actually a map into the unit circle. By previous exercises, it suffices to show that the image is compact or that the homomorphism is continuous. I found a similar exercise in Rudin's Real and Complex Analysis which asks to show that every Lebesgue measurable character on $\mathbb{R}$ is continuous. After spending a good deal of time playing around with the solution to that exercise, I realized that Rudin defines a character as a complex homomorphism having modulus 1, presupposing the result in a fundamental way.
I succeeded in showing (by copying the proof of Rudin's Theorem 9.23 and the exercise), given that a measurable homomorphism must have modulus 1, it must also be continuous and therefore given by an exponential. This is a later problem in Katznelson's book. A hint to the original problem would be appreciated!

REPLY [7 votes]: Here's a proof that doesn't use any tools. Let $\chi:\mathbb{T}\to\mathbb{C}^{\times}$ be a measurable homomorphism. Consider the sets $C_n=\{z\in\mathbb{C}\mid 1/n<|z|<n\}$. By countable subadditivity of measures, there exists $n$ such that the set $B_n=\chi^{-1}(C_n)$ has measure at least $9/10$ (where $\mathbb{T}$ has measure $1$).
If there exists $z\in\mathbb{T}$ such that $|\chi(z)|\neq 1$, we can pick an integer $k$ such that $|\chi(z)|^k\geq n^2$. So $\chi(z)^k C_n$ is disjoint from $C_n$. But the measure on $\mathbb{T}$ is preserved by multiplication, so $z^k B_n$ has the same measure as $B_n$. This implies $9/10+9/10\leq 1$, a contradiction.