Measurable homomorphism of $\mathbb{T}$ into $\mathbb{C}^\times$ 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!
 A: 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.
A: I'm going to type up an answer. I'll leave it up for a few days for feedback or alternate answers before accepting.
Let $\chi:\mathbb{T}\rightarrow\mathbb{C}^\times$ be a measurable homomorphism; it suffices to show that $\chi$ bounded. Suppose $|\chi|\leq C$ and  there exists a $t\in\mathbb{T}$ such that $|\chi(t)|\neq1$. Then either $|\chi(t)|$ or $|\chi(-t)|$ is greater than 1 and so $|\chi(t)|^n$ or $|\chi(-t)|^n$ is eventually greater than $C$. Hence if $\chi$ is bounded, then $|\chi(t)|=1$ for every $t$ in $\mathbb{T}$.
We now show $\chi$ must be bounded. Because $\chi$ is measurable, Luzin's theorem provides the existence of a continuous $f$ such that $\chi$ agrees with $f$ on a subset $E$ of $\mathbb{T}$ of positive measure. As $\mathbb{T}$ is compact, $|f|$ is bounded by some $C>1$, hence $\chi$ is bounded on $E$. The Steinhaus Theorem shows that $E$ generates $\mathbb{T}$ as an additive group and in fact, as $\mathbb{T}$ is bounded, there exists a positive integer $N$ such that $E+E+\cdots+E=\mathbb{T}$ where the sum has $N$ terms. (This was actually a previous exercise in Katznelson.)
Taking an arbitrary $t$ in $\mathbb{T}$, write it as $t=t_1+t_2+\cdots+t_k$ where $t_i \in E$ for $i=1,2,\dots,k \leq N$. As the $t_i$ are contained in $E$, we know that $|\chi(t_i)|\leq C$. Because $\chi$ is a homomorphism,
$$|\chi(t)|=|\chi(t_1+\cdots+t_k)|=|\chi(t_1)|\cdots|\chi(t_k)|\leq C^N.$$
Therefore $\chi$ is bounded on $\mathbb{T}$, completing the proof.
