## Introduction

In the study of measures, besides understanding when one measure is controlled by another (absolute [continuity](/page/Continuity)), it's equally important to have a notion for when two measures are entirely disjoint. This is captured by the concept of **mutual singularity**. Two measures, $\nu$ and $\mu$, are mutually singular if they "live" on separate, disjoint [sets](/page/Set). More formally, we can find a set $B$ such that all of the "mass" of $\nu$ is concentrated inside $B$, while all of the "mass" of $\mu$ is concentrated outside of $B$.

A classic example is the relationship between the Lebesgue measure on the real line and a point mass measure. Let $\mu = \mathcal{L}^1$ be the Lebesgue measure, and let $\nu = \delta_0$ be the Dirac measure at zero, where $\nu(A) = 1$ if $0 \in A$ and $\nu(A) = 0$ if $0 \notin A$. If we choose the set $B = \{0\}$, then $\nu(\mathbb{R} \setminus B) = \nu(\mathbb{R} \setminus \{0\}) = 0$ and $\mu(B) = \mu(\{0\}) = 0$. Thus, these two measures are mutually singular. This concept allows us to formalize the separation of different types of measure-theoretic behaviors.

## Formal Definition and Key Results

We begin with the formal definition of mutual singularity.

[definition] Let $\mu$ and $\nu$ be two $\sigma$-finite Borel measures on $\mathbb{R}^n$. We say that $\nu$ and $\mu$ are **mutually singular**, written $\nu \perp \mu$, if there exists a Borel set $B \subseteq \mathbb{R}^n$ such that $\mu(B) = 0$ and $\nu(\mathbb{R}^n \setminus B) = 0$. The set $B$ is sometimes called a separating set. [/definition]

[remark] The condition is symmetric. If $\nu \perp \mu$, there is a set $B$ with $\mu(B) = 0$ and $\nu(\mathbb{R}^n \setminus B) = 0$. Let $A = \mathbb{R}^n \setminus B$. Then $\nu(A) = \nu(\mathbb{R}^n \setminus B) = 0$. Also, $\mu(\mathbb{R}^n \setminus A) = \mu(B) = 0$. This shows that the definition is symmetric, so $\nu \perp \mu$ is equivalent to $\mu \perp \nu$. [/remark]

The most important result related to mutual singularity is the Lebesgue Decomposition Theorem, which shows that any $\sigma$-finite measure can be split into an absolutely continuous part and a mutually singular part with respect to another $\sigma$-finite measure.

[theorem] **(Lebesgue Decomposition Theorem)** Let $\mu$ and $\nu$ be $\sigma$-finite Borel measures on $\mathbb{R}^n$. Then there exists a unique pair of measures, $\nu_{ac}$ and $\nu_s$, such that: 1. $\nu = \nu_{ac} + \nu_s$. 2. $\nu_{ac} \ll \mu$ (absolutely continuous with respect to $\mu$). 3. $\nu_s \perp \mu$ (singular with respect to $\mu$). Furthermore, by the Radon-Nikodym theorem, there exists a non-negative function $f \in L^1(\mu)$ such that $d\nu_{ac} = f \, d\mu$. Thus, the decomposition can be written as: \begin{align*} \nu(A) = \int_A f \, d\mu + \nu_s(A) \end{align*} for any Borel set $A \subseteq \mathbb{R}^n$. [/theorem]

This theorem provides a complete structural description of how any two $\sigma$-finite measures relate. It tells us that given a reference measure $\mu$, any other measure $\nu$ can be broken down into two components: a part $\nu_{ac}$ that "behaves like" $\mu$ (in the sense that it has a density with respect to $\mu$) and a part $\nu_s$ that is completely "foreign" to $\mu$. This decomposition is powerful because it allows us to study these two different behaviors separately. It is the foundation for understanding mixed-type probability [distributions](/page/Distribution) and for separating regular and singular phenomena in analysis and PDEs.

We now prove the uniqueness of this decomposition.

[proof] **Proof of Uniqueness** Suppose we have two such decompositions of $\nu$: \begin{align*} \nu = \nu_{ac} + \nu_s \quad \text{and} \quad \nu = \nu'_{ac} + \nu'_s \end{align*} where $\nu_{ac} \ll \mu$, $\nu'_{ac} \ll \mu$, $\nu_s \perp \mu$, and $\nu'_s \perp \mu$. Our goal is to show that $\nu_{ac} = \nu'_{ac}$ and $\nu_s = \nu'_s$. **Step 1 (Equate the decompositions)** From the two decompositions, we can write: \begin{align*} \nu_{ac} + \nu_s = \nu'_{ac} + \nu'_s \end{align*} Rearranging this equation gives: \begin{align*} \nu_{ac} - \nu'_{ac} = \nu'_s - \nu_s \end{align*} Let's define a signed measure $\lambda = \nu_{ac} - \nu'_{ac}$. Our task is to show that $\lambda$ is the zero measure. **Step 2 (Show that λ is absolutely continuous with respect to μ)** Let $A$ be a Borel set such that $\mu(A) = 0$. Since $\nu_{ac} \ll \mu$, we have $\nu_{ac}(A) = 0$. Similarly, since $\nu'_{ac} \ll \mu$, we have $\nu'_{ac}(A) = 0$. The Jordan decomposition for signed measures allows us to state that if the positive and negative parts of the measure are zero on A, then the measure of A is zero. Therefore, \begin{align*} \lambda(A) = \nu_{ac}(A) - \nu'_{ac}(A) = 0 - 0 = 0. \end{align*} This holds for any set $A$ with $\mu(A) = 0$, so we conclude that $\lambda \ll \mu$. **Step 3 (Show that λ is singular with respect to μ)** Since $\nu_s \perp \mu$, there exists a set $B_1$ such that $\mu(B_1)=0$ and $\nu_s(\mathbb{R}^n \setminus B_1)=0$. Since $\nu'_s \perp \mu$, there exists a set $B_2$ such that $\mu(B_2)=0$ and $\nu'_s(\mathbb{R}^n \setminus B_2)=0$. Let $B = B_1 \cup B_2$. By the sub-additivity of measures, $\mu(B) \le \mu(B_1) + \mu(B_2) = 0+0=0$. So, $\mu(B)=0$. Now consider the complement, $\mathbb{R}^n \setminus B = (\mathbb{R}^n \setminus B_1) \cap (\mathbb{R}^n \setminus B_2)$. For any measurable set $E \subseteq \mathbb{R}^n \setminus B$, we have $E \subseteq \mathbb{R}^n \setminus B_1$, which implies $\nu_s(E) = 0$. Similarly, $E \subseteq \mathbb{R}^n \setminus B_2$, which implies $\nu'_s(E)=0$. This shows that both $\nu_s$ and $\nu'_s$ are concentrated on $B$. Therefore, their difference, $\lambda = \nu'_s - \nu_s$, must also be concentrated on $B$. Formally, for any set $E \subseteq \mathbb{R}^n \setminus B$, $\lambda(E) = \nu'_s(E) - \nu_s(E) = 0-0=0$. This means $\lambda(\mathbb{R}^n \setminus B)=0$. Since we have found a set $B$ with $\mu(B) = 0$ and $\lambda(\mathbb{R}^n \setminus B) = 0$, we conclude that $\lambda \perp \mu$. **Step 4 (Conclude λ is the zero measure)** We have shown that the signed measure $\lambda$ is both absolutely continuous with respect to $\mu$ and singular with respect to $\mu$. Since $\lambda \perp \mu$, there is a set $B$ with $\mu(B) = 0$ and $\lambda(\mathbb{R}^n \setminus B) = 0$. Since $\lambda \ll \mu$ and $\mu(B) = 0$, we must have $\lambda(B) = 0$. The total measure of the entire space is \begin{align*} \lambda(\mathbb{R}^n) = \lambda(B) + \lambda(\mathbb{R}^n \setminus B) = 0 + 0 = 0. \end{align*} A similar argument for the positive and negative parts in the Jordan decomposition of $\lambda$ shows that $\lambda$ must be the zero measure. That is, $\lambda(A) = 0$ for all Borel sets $A$. **Step 5 (Establish uniqueness)** Since $\lambda = 0$, we have $\nu_{ac} - \nu'_{ac} = 0$, which implies $\nu_{ac} = \nu'_{ac}$. It follows immediately that $\nu'_s - \nu_s = 0$, which implies $\nu_s = \nu'_s$. This completes the proof of uniqueness. [/proof]

## Applications

The Lebesgue Decomposition Theorem is a fundamental structural tool with applications throughout mathematics.

### Probability Theory: Decomposing Random Variables

The distribution of any random variable can be classified using the Lebesgue decomposition. Let $X$ be a real-valued random variable with distribution measure $\nu$. We can decompose $\nu$ with respect to the Lebesgue measure $\mathcal{L}^1$ on $\mathbb{R}$.

\begin{align*} \nu = \nu_{ac} + \nu_s \end{align*}

* The absolutely continuous part, $\nu_{ac}$, corresponds to the part of the distribution that has a **probability density function (PDF)**. Its Radon-Nikodym [derivative](/page/Derivative) $f = \frac{d\nu_{ac}}{d\mathcal{L}^1}$ is the PDF. * The singular part, $\nu_s$, corresponds to the part of the distribution concentrated on a set of Lebesgue measure zero. This singular part can itself be decomposed into a **discrete part** (a sum of Dirac masses at specific points, representing a probability mass function) and a **singular continuous part** (a measure concentrated on an uncountable set of measure zero, like the Cantor set).

**Example:** Consider a process where a coin is flipped. If heads, the outcome is 0. If tails, the outcome is a number drawn uniformly from $[2, 3]$. The distribution measure $\nu$ of the outcome is a **mixed distribution**. Its Lebesgue decomposition with respect to $\mathcal{L}^1$ is:

* $\nu_s = \frac{1}{2} \delta_0$. This is a point mass at 0, which is singular to $\mathcal{L}^1$. * $d\nu_{ac} = f(x) \, dx$, where $f(x) = \frac{1}{2}$ for $x \in [2, 3]$ and $0$ otherwise. This part is absolutely continuous.