The [Fourier transform](/page/Fourier%20Transform) assigns to each [integrable](/page/Integral) function $f \in L^1(\mathbb{R}^n, \mathcal{L}^n)$ a continuous function $\hat{f}$ on frequency space. But many of the most important objects in analysis and probability — point masses, singular distributions of random variables, spectral measures of stochastic processes — are not [functions](/page/Function) at all. They are measures, and the integral defining $\hat{f}$ makes no sense for them because there is no density to integrate against Lebesgue measure. The Fourier–Stieltjes transform resolves this by replacing $f(x) \, d\mathcal{L}^n(x)$ with a general finite Borel measure $\mu$ in the exponent-weighted integral, producing a bounded continuous function that encodes the oscillatory content of $\mu$ in the same way that $\hat{f}$ encodes the oscillatory content of $f$. In probability, this object is the characteristic function of a random variable; in harmonic analysis, it is the natural extension of the Fourier transform to the measure algebra $M(\mathbb{R}^n)$.

## Motivation

[motivation] ### The $L^1$ [Boundary](/page/Boundary) The [Fourier transform](/page/Fourier%20Transform) on $L^1(\mathbb{R}^n, \mathcal{L}^n)$ is defined by \begin{align*} \hat{f}(\xi) &= \int_{\mathbb{R}^n} f(x) \, e^{i\xi \cdot x} \, d\mathcal{L}^n(x). \end{align*} This integral converges absolutely because $|f(x) e^{i\xi \cdot x}| = |f(x)|$ and $f \in L^1$. The [Riemann–Lebesgue lemma](/theorems/245) guarantees that $\hat{f}$ is bounded, uniformly continuous, and vanishes at infinity: $\hat{f} \in C_0(\mathbb{R}^n)$. These are powerful properties, and within $L^1$ the theory works well. The difficulty is that $L^1$ is not large enough. Consider the Dirac mass $\delta_0$, defined by $\delta_0(A) = \mathbb{1}_A(0)$ for every Borel set $A \subseteq \mathbb{R}^n$. This is a perfectly well-defined finite Borel measure — it assigns mass $1$ to any set containing the origin — but it has no density with respect to Lebesgue measure. There is no function $f \in L^1(\mathbb{R}^n, \mathcal{L}^n)$ such that $\delta_0(A) = \int_A f \, d\mathcal{L}^n$ for all Borel sets $A$. The $L^1$ Fourier transform therefore cannot see $\delta_0$ at all. Yet the formal computation is immediate: if we insert $\delta_0$ into the integral, we obtain \begin{align*} \int_{\mathbb{R}^n} e^{i\xi \cdot x} \, d\delta_0(x) &= e^{i\xi \cdot 0} = 1. \end{align*} The "Fourier transform" of the Dirac mass is the constant function $1$ — a perfectly concrete object. The integral makes sense; it is only the insistence on having a Lebesgue density that fails. This suggests that the correct domain for the Fourier transform should include measures, not just integrable functions. ### Measures as the Natural Domain An $L^1$ function $f$ acts on frequency space through the integral $\xi \mapsto \int f(x) e^{i\xi \cdot x} \, d\mathcal{L}^n(x)$. But this integral depends on $f$ only through the measure $\mu_f$ defined by $\mu_f(A) = \int_A f \, d\mathcal{L}^n$ — the absolutely continuous measure with density $f$. The mapping $f \mapsto \mu_f$ embeds $L^1(\mathbb{R}^n, \mathcal{L}^n)$ isometrically into the space $M(\mathbb{R}^n)$ of finite complex Borel measures (equipped with the total variation norm), and the Fourier transform of $f$ is really the Fourier transform of the measure $\mu_f$: \begin{align*} \hat{f}(\xi) &= \int_{\mathbb{R}^n} e^{i\xi \cdot x} \, d\mu_f(x). \end{align*} Once this is recognised, the generalisation is immediate: for any $\mu \in M(\mathbb{R}^n)$, the integral $\int e^{i\xi \cdot x} \, d\mu(x)$ converges absolutely because $|e^{i\xi \cdot x}| = 1$ and $|\mu|(\mathbb{R}^n) < \infty$. No density is required, and the resulting function of $\xi$ inherits [continuity](/page/Continuity) and boundedness from the [dominated convergence theorem](/theorems/4). The space $M(\mathbb{R}^n)$ is strictly larger than $L^1$: it contains point masses, singular continuous measures (such as the Cantor measure), and all convex combinations thereof. ### The Probabilistic Perspective In probability theory, the same object appears under a different name. If $X: \Omega \to \mathbb{R}^n$ is a random variable on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$, its law $\mu_X = \mathbb{P} \circ X^{-1}$ is a Borel probability measure on $\mathbb{R}^n$, and the characteristic function of $X$ is \begin{align*} \phi_X(\xi) &= \mathbb{E}[e^{i\xi \cdot X}] = \int_{\mathbb{R}^n} e^{i\xi \cdot x} \, d\mu_X(x). \end{align*} This is exactly the Fourier–Stieltjes transform of $\mu_X$. The characteristic function determines the distribution uniquely (the uniqueness theorem below), encodes all moments through its Taylor expansion at the origin, and characterises convergence in distribution through the Lévy continuity theorem. These facts make it the single most important analytic tool in probability theory — and they are all consequences of the general theory of the Fourier–Stieltjes transform. [/motivation]

## The Fourier–Stieltjes Transform

### The Stieltjes Connection: Measures and Bounded Variation

The name "Stieltjes" refers to the Riemann–Stieltjes integral: the classical formulation defines the transform by integrating $e^{i\xi x}$ not against a density $f(x) \, d\mathcal{L}^1(x)$, but against a function of [bounded variation](/page/Functions%20of%20Bounded%20Variation) $dF(x)$. In one dimension, every finite Borel measure $\mu$ on $\mathbb{R}$ corresponds to a right-continuous BV function $F: \mathbb{R} \to \mathbb{C}$ via the **distribution function**

\begin{align*} F(x) := \mu((-\infty, x]), \end{align*}

and the Fourier–Stieltjes transform of $\mu$ coincides with the Riemann–Stieltjes integral:

\begin{align*} \hat{\mu}(\xi) = \int_{-\infty}^\infty e^{i\xi x} \, dF(x). \end{align*}

The total variation of $\mu$ as a measure equals the total variation of $F$ as a BV function: $|\mu|(\mathbb{R}) = V(F; \mathbb{R})$. This correspondence is a bijection: every right-continuous BV function $F$ with $\lim_{x \to -\infty} F(x) = 0$ determines a unique finite Borel measure, and conversely.

[example: Distribution Function Of A Mixed Measure] Let $\mu = \frac{1}{2}\mathcal{L}^1|_{[0,1]} + \frac{1}{2}\delta_2$ — half of its mass is spread uniformly over $[0,1]$, and the other half is a point mass at $x = 2$. The distribution function is: \begin{align*} F(x) = \begin{cases} 0 & \text{if } x < 0, \\ x/2 & \text{if } 0 \leq x < 1, \\ 1/2 & \text{if } 1 \leq x < 2, \\ 1 & \text{if } x \geq 2. \end{cases} \end{align*} The function $F$ is right-continuous, non-decreasing, with a continuous part (the ramp on $[0,1]$) and a jump of size $1/2$ at $x = 2$. Its total variation is $V(F; \mathbb{R}) = 1 = |\mu|(\mathbb{R})$. The Fourier–Stieltjes transform of $\mu$ can be computed as the Riemann–Stieltjes integral $\int e^{i\xi x} \, dF(x)$, which splits into the continuous part $\frac{1}{2}\int_0^1 e^{i\xi x} \, d\mathcal{L}^1(x)$ and the jump contribution $\frac{1}{2}e^{2i\xi}$. [/example]

In higher dimensions ($n \geq 2$), there is no natural analogue of distribution functions, and the measure-theoretic formulation becomes essential. But in $\mathbb{R}^1$, the BV and measure viewpoints are interchangeable, and the BV formulation is how the transform appears in classical probability and in older texts on harmonic analysis.

### The Measure-Theoretic Definition

To state the definition in its modern form, we need the ambient space of finite measures. Without this space, we have no norm to control the transform and no algebraic structure ([convolution](/page/Convolution)) to exploit. The space of finite complex Borel measures is to the Fourier–Stieltjes transform what $L^1(\mathbb{R}^n)$ is to the ordinary Fourier transform: the natural domain on which the transform is a bounded [linear map](/page/Linear%20Map).

[definition:Space Of Finite Complex Borel Measures] Let $M(\mathbb{R}^n)$ denote the space of all finite complex Borel measures on $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$: that is, the space of all countably additive set functions $\mu: \mathcal{B}(\mathbb{R}^n) \to \mathbb{C}$ with finite total variation \begin{align*} \|\mu\| &:= |\mu|(\mathbb{R}^n) < \infty, \end{align*} where $|\mu|$ is the total variation measure of $\mu$. Equipped with the total variation norm $\|\cdot\|$ and the convolution product (defined below), $M(\mathbb{R}^n)$ is a unital Banach algebra with identity $\delta_0$. [/definition]

The total variation norm is the natural norm on $M(\mathbb{R}^n)$: it measures the "total mass" of $\mu$ when sign and phase are ignored. For a positive measure, the total variation equals the total mass $\mu(\mathbb{R}^n)$; for a signed or complex measure, it accounts for cancellation.

With this space in hand, the Fourier–Stieltjes transform is defined by the same exponential integral that defines the $L^1$ Fourier transform, but with the Lebesgue density replaced by a general measure.

[definition:Fourier Stieltjes Transform] Let $\mu \in M(\mathbb{R}^n)$. The **Fourier–Stieltjes transform** of $\mu$ is the function \begin{align*} \hat{\mu}: \mathbb{R}^n &\to \mathbb{C} \\ \xi &\mapsto \int_{\mathbb{R}^n} e^{i\xi \cdot x} \, d\mu(x). \end{align*} [/definition]

The sign convention $e^{i\xi \cdot x}$ (rather than $e^{-i\xi \cdot x}$) matches the standard probabilistic convention for characteristic functions. The analysis convention, used on the [Fourier Transform](/page/Fourier%20Transform) page, differs by a sign in the exponent; when $\mu$ has a density $f \in L^1$, the two conventions are related by $\hat{\mu}(\xi) = \hat{f}(-\xi)$. Either choice leads to an equivalent theory, but one must be consistent. Throughout this page, we use $e^{i\xi \cdot x}$.

The integral converges absolutely for every $\xi \in \mathbb{R}^n$ because