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

$||(f-g) \chi_{[-K, K]^{d}} ||_{\infty} < \epsilon$


$||(f-g)\chi_{\mathbb{R}^{d} - [-K, K]^{d}} ||_{\infty} \leq ||g||_{\infty} = ||\tilde{g}||_{\infty} \leq ||\tilde{f}||_{\infty} + \epsilon = ||f||_{\infty} + \epsilon$.

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!

  • 1
    $\begingroup$ @OliverDíaz Thanks, my first time posting a question. I've typed the theorem from the book $\endgroup$
    – Tom
    Commented Oct 31, 2022 at 20:57
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    $\begingroup$ Here is a posting related to the inversion formula of the Fourier transform of finite measures. The point is that you can recover the measure of rectangles from the characteristic function. If two measures coincide on rectangles, then they are the same. $\endgroup$
    – Mittens
    Commented Nov 2, 2022 at 0:57

1 Answer 1


The key idea lurking in this proof is that $\tilde f$ should be the periodically extended version of $f$. If for example $f$ is supported inside the unit cube, then we can tile space with such cubes and define $\tilde f$ to look like $f$ inside each of these cubes. A rigorous way to define $\tilde f$ is by summing all translates of $f$ using all integer multiples of the fundamental lattice periods. That is, if $f$ is supported in the unit cube then $\tilde f(x)= \sum_K f(x+K)$ where the summation extends over all $K\in {\mathbb Z}^n$. (The author seems to have left out the summation sign in their construction of $\tilde f$.)


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