When does the integral over points on the unit circle get 1 for a probability measure?

I am studying probability theory and want to solve the following exercise.

Exercise. Let $$X$$ be a real random variable. Then the following are equal:

• $$\mathbb{P}\left( X \in a + b\mathbb{Z}\right) = 1$$ for some $$a, b \in \mathbb{R}$$.
• There exists a $$z \in \mathbb{R}$$ such that $$\lvert \mathbb{\hat{P}}\left(z\right)\rvert = 1$$.

Here, $$\mathbb{\hat{P}}$$ is the characteristic function of the probability measure $$\mathbb{P}$$.

I was able to show "$$\Rightarrow$$" but for the other direction I guess I need this statement:

For any probability measure $$\mathbb{P}$$, if we have

$$\left\lvert \int_\mathbb{R} e^{itx} d\mathbb{P}\left(x\right) \right\rvert = 1 \;\text{for a}\; t \in\mathbb{R}\setminus\{0\},$$

then $$\exp(itx) = u$$ for a $$u \in \mathbb{C}$$ with $$\lvert u \rvert = 1$$ (almost surely).

This is clear to me from graphical considerations but I have no clue how to show it formally.

• is the first bullet supposed to be "for some $a,b \in \mathbb{R}$"? – mathworker21 Nov 29 '18 at 17:43
• @mathworker21Yes. I'll change that. – fpmoo Nov 29 '18 at 17:45

The characteristic function's square modulus is $$(\mathbb{E}\cos tX)^2+(\mathbb{E}\sin tX)^2=1-\operatorname{Var}\cos tX-\operatorname{Var}\sin tX$$, requiring $$\cos tX,\,\sin tX$$ to be constant variables, which is equivalent to $$\tan\tfrac{tX}{2}$$ being constant.
You wish to show that if $$f:\mathbb{R} \to \mathbb{C}$$ is s.t. $$|f(x)| \le 1$$ for each $$x$$ and $$|\int_\mathbb{R} f(x)d\mu(x)| = 1$$ for some probability measure $$\mu$$, then $$f$$ is constant $$\mu$$-a.e..
By multiplying $$f$$ by a constant, we may assume $$\int_\mathbb{R} f(x)d\mu(x) = 1$$. Then, $$\int_\mathbb{R} Re[f(x)]d\mu(x) = 1$$ and $$Re[f(x)] \le 1$$ for each $$x$$, so we must have $$Re[f(x)] = 1$$ for $$\mu$$-a.e. $$x$$. It follows that $$f = 1$$ $$\mu$$-a.e..