The theorem 15.9 (Characteristic function) says: A finite measure $\mu \in \mathcal{M}_{f}(\mathbb{R}^{d})$ (here this symbol just means finite measures on $\mathbb{R}^{d}$) is characterized by its characteristic function.
Proof: Let $\mu_{1}, \mu_{2} \in \mathcal{M}_{f}(\mathbb{R}^{d})$ with $\psi_{\mu_{1}}(t) = \psi_{\mu_{2}}(t)$ for all $t \in \mathbb{R}^{d}$. By Theorem 13.11(ii), $C_{c}(\mathbb{R}^{d})$ is a separating class for $\mathcal{M}_{f}(\mathbb{R}^{d})$. Hence it is enough to show that $\int f d\mu_{1} = \int f d\mu_{2}$ for all $f \in C_{c}(\mathbb{R}^{d})$.
Let $f: \mathbb{R}^{d} \xrightarrow{} \mathbb{R}$ be continuous with compact support and let $\epsilon > 0$. Assume that $K > 0$ is large enough such that $f(x) = 0$ for $x \notin (-K/2, K/2)^{d}$ and such that $\mu_{i}(\mathbb{R}^{d} - (-K, K)^{d}) < \epsilon, i = 1, 2$. Consider the torus $E := \mathbb{R}^{d} / (2K\mathbb{Z}^{d})$ and define $\tilde{f} : E \xrightarrow{} \mathbb{R}$ by
$\tilde{f}(x + 2K\mathbb{Z}^{d}) = f(x)$ for $x \in [-K, K)^{d}$.
Since the support of $f$ is contained in $(-K, K)^{d}$, $\tilde{f}$ is continuous.
For $m \in \mathbb{Z}^{d}$ define
$g_{m}: \mathbb{R}^{d} \xrightarrow{} \mathbb{C}, x \mapsto \text{exp}(i \langle \pi m / K, x \rangle)$.
Let $\mathcal{C}$ be the algebra of finite linear combinations of the $g_{m}$. For $g \in \mathcal{C}$, we have $g(x) = g(x + 2Kn)$ for all $x \in \mathbb{R}^{d}$ and $n \in \mathbb{Z}^{d}$. Hence, the map
$\tilde{g} : E \xrightarrow{} \mathbb{C}, \tilde{g}(x + 2K \mathbb{Z}^{d}) = g(x)$
is well-defined, continuous and bounded. Furthermore, $\tilde{\mathcal{C}} := \{\tilde{g} : g \in \mathcal{C}\} \subset C_{b}(E; \mathbb{C})$ is an algebra that separates points and is closed under complex conjugation. As $E$ is compact, by the Stone-Weierstrass theorem, there is a $g \in \mathcal{C}$ such that $||\tilde{g} - \tilde{f} ||_{\infty} < \epsilon$. We infer
$||(f-g) \chi_{[-K, K]^{d}} ||_{\infty} < \epsilon$
and
$||(f-g)\chi_{\mathbb{R}^{d} - [-K, K]^{d}} ||_{\infty} \leq ||g||_{\infty} = ||\tilde{g}||_{\infty} \leq ||\tilde{f}||_{\infty} + \epsilon = ||f||_{\infty} + \epsilon$.
By assumption of the theorem, $\int g d \mu_{1} = \int g d \mu_{2}$. Hence, using the triangle inequality, we conclude
$|\int f d \mu_{1} - \int f d \mu_{2}| \leq \int |f - g| d \mu_{1} + \int |f - g| d \mu_{2} \leq \epsilon(2 ||f ||_{\infty} + 2\epsilon + \mu_{1}(\mathbb{R}^{d}) + \mu_{2}(\mathbb{R}^{d}))$.
As $\epsilon$ is arbitrary, the integrals coincide.
Why is this $\tilde{f}$ well defined? That is, if $x, y$ are from the same left coset, i.e., $x + 2K\mathbb{Z}^{d} = y + 2K\mathbb{Z}^{d}$, how to see $f(x) = f(y)$?
I found here https://proofwiki.org/wiki/Elements_in_Same_Left_Coset_iff_Product_with_Inverse_in_Subgroup#:~:text=Let%20x%2Cy%E2%88%88G,x%E2%88%921y%E2%88%88H. that $x, y$ are in the same left coset of $H$ iff $x^{-1}y$ is in the $H$.
For any $z$ outside $[-K, K)^{d}$, $z$ is equivalent to some $x \in [-K, K)^{d}$, with $z + 2K\mathbb{Z}^{d} = x + 2K\mathbb{Z}^{d}$ and so by definition $\tilde{f}(z + 2K\mathbb{Z}^{d}) = \tilde{f}(x + 2K\mathbb{Z}^{d}) = f(x)$. So to see it is well-defined it's enough to consider two points $x, y \in [-K, K)^{d}$.
If $x + 2K\mathbb{Z}^{d} = y + 2K\mathbb{Z}^{d}$, then $y - x = 2Km$, for some $m \in \mathbb{Z}^{d}$. And recall that $\tilde{f}(x + 2K\mathbb{Z}^{d})$ is defined only for $x \in [-K, K)^{d}$, and for any two vectors from $[-K, K)^{d}$, their difference in each coordinate is less than $2K$, so such $m$ must be zero, meaning $x = y$.
My question: What does it mean for $\tilde{f}$ to be continuous? Why does $f$'s being supported within $(-K, K)^{d}$ imply $\tilde{f}$ is continuous?
Here is the definition of the quotient space: https://en.wikipedia.org/wiki/Quotient_space_(topology). Given a topological space $X$, the quotient space under ~ is the quotient set $Y$ equipped with the quotient topology, which is the topology whose open sets are the subsets $U \subset Y = X / \text{~}$ such that $\{x \in X: [x] \in U\}$ is an open subset of $X$; that is, $U \subset X / \text{~}$ iff $\{x \in X: [x] \in U\} \in \tau_{X}$. Now $Y = \{x + 2K\mathbb{Z}^{d}: x \in [-K, K)^{d}\}$ so any subset $U$ of $Y$ has the form $U = \{x_{i} + 2K\mathbb{Z}^{d}\}_{i \in I}$. If such $U$ is open in the quotient topology, this means $\{x_{i}\}_{i \in I}$ is open in $X$.
For any open interval of the form $A = (a, b) \in \mathbb{R}$, we want to show $(\tilde{f})^{-1}(A)$ is open in the quotient topology. Since $f$ is continuous, $f^{-1}(A)$ is open in $\mathbb{R}^{d}$. What next? (I don't know if this works: Denote $A = \{x_{j}\}_{j \in J}$. Then $\{[x_{j}]\}_{j \in J}$ is open in the quotient topology. Since for any $z \in \mathbb{R}^{d}$, $z$ is in the same equivalence class as $z'$ for some $z' \in [-K, K)^{d}$, we can rewrite $\{[x_{j}]\}_{j \in J} = \{[x_{t}]\}_{t \in T}$ where each $x_{t} \in [-K, K)^{d}$. By the definition of quotient topology, that $\{[x_{j}]\}_{j \in J} = \{[x_{t}]\}_{t \in T}$ is open implies $\{x_{t}\}_{t \in T}$ is open. )
This theorem is really scary to me (I learned measure theory mainly from the first five chapters from Donald Cohn's book, and I really like his style where every step is explained clearly. So when the author just omits something when they could have used some more words to explain a bit further it causes difficulty to my understanding. Maybe the details missing here are obvious to you but I really can't understand)... Appreciate any help!
