Your approach using the spectral theorem will succeed. A further hint is to split $\nu$ into components $\nu_0+\nu_1$, where $\nu_0$ is a sum of atomic measures (i.e. $\nu_0 = \sum_{n} c_n\delta_{x_n}$) and $\nu_1$ is continuous, meaning $\nu_1(\{x\})=0$ for all $x.$

@JohnGriesmer I thought about your comment. The nonzero eigenfunctions are precisely the functions which are supported at a single point of positive measure. So if the continuous part of $\nu$ is $0$ then $1$ can be written as a sum of eigenfunctions. So I guess almost periodicity of $1$ is equivalent to the fact that the continuous part of $\nu$ vanishes (though I do not see how to prove it yet). Am I right?

You are correct.

@JohnGriesmer I am stuck. Assuming $1$ is almost periodic, I get that the set $\{\chi_n:\ n\in \mathbb Z\}$ is precompact in $L^2(\mathbb T, \nu)$, where $\chi_n$ is the $n$-th character. I am not able to see how I can use this. Can you please help.

You want to prove that if $\{\chi_n:n\in \mathbb Z\}$ is precompact, then $\nu$ is purely atomic. Wiener's lemma (Wikipedia) will help you prove the contrapositive: if $\nu$ is not purely atomic, then the set of characters is not precompact. It's probably easiest to work this out by first proving the following special case: if $\nu$ is continuous, then the set of characters is not precompact. Also Wiener's lemma has a very nice proof which is worthwhile on its own, if you've never seen it before.

@JohnGriesmer So let $\nu$ be a continuous probability measure on $\mathbb T$ and I want to show that the characters cannot form a precompact set. Using Wiener's lemma, the coninuity of $\nu$ yeilds that $\hat \nu(n)\to 0$ as $n\to \infty$ away from a density $0$ set. But I am not getting anywhere.

Sorry, I didn't realize moving to chat would unto the LaTeX.

If it's okay with you I can digest the non TeX version.

Assuming that $\{\chi_n:n\in \mathbb Z\}$ is precompact, then for all $\epsilon>0$, there are finitely many characters $\{\chi_{n_1},\dots,\chi_{n_k}\}$ such that for all $n\in \mathbb N$, there is a $\chi_{n_j}$ such that $\langle \chi_n , \chi_{n_j} \rangle_{L^2(\nu)} > 1-\epsilon$. Thus, $\sum_{j=1}^k \frac{1}{2N+1}\sum_{n= -N}^N \langle \chi_n, \chi_{n_j}\rangle>\epsilon.$ (cont)

for every $N$.

On the other hand, Wiener's lemma will tell you that the latter sum converges to 0 as $N\to \infty$.

Sorry, $\sum_{j=1}^k \frac{1}{2N+1}\sum_{n= -N}^N \langle \chi_n, \chi_{n_j}\rangle>\epsilon.$ should be

$\sum_{j=1}^k \frac{1}{2N+1}\sum_{n= -N}^N \langle \chi_n, \chi_{n_j}\rangle > 1-\epsilon.$

Anyway, if this chat business isn't working I'll post a complete answer on the question page.

Also, I've been meaning to answer your other question about projection onto the Kronecker factor.

I actually had a question about that: when you say "Birkhoff type theorem," do you mean pointwise almost-everywhere convergence? Because it looks like you're using "Birkhoff" to refer to convergence in the strong operator topology, and I always have seen "Birkhoff" refer to pointwise almost-everywhere convergence.

$\sum_{j=1}^k \frac{1}{2N+1}\sum_{n= -N}^N |\langle \chi_n, \chi_{n_j}\rangle| > 1-\epsilon.$

I had previously written 'Birkhoff' where I should have written 'von Neumann.'

I had changed it I think.

I am reading your comment. Will take some time.

I see. Either way it's a very interesting question. Asking for the Birkhoff version will lead to the Wiener-Wintner theorem.

Good to see someone finds it interesting!

The comments by @mathworker kind of discouraged me.

If you can write an answer then that will be great. That way I can accept your answer too.

LoL. Well it's clear from your post history that you're a serious mathematician. Don't be discouraged by random commenters. Or even big shots; if you happen to get a negative comment from one of them.

I'll write the answer now.

Thanks. It means a lot.

I see your main idea in your proof now.

Well I put up an answer. For some reason I thought you were trying to prove Tao's Proposition 2, but now I posted an answer to your actual question about Exercise 5.

But this spectral theory/Herglotz lemma/Wiener's lemma stuff is very useful to think about.

Thank you for the answer. I will take some time to accept. Also, thanks for letting me know about Wiener's lemma and its application.