Uniqueness of the Lebesgue Decomposition

What Uniqueness Says

A Lebesgue decomposition separates a measure into a part that vanishes on sets of reference measure zero and a part concentrated on a reference-null set. The existence of such a split and its uniqueness are different questions: once a decomposition is given, uniqueness says that no alternative allocation of mass can satisfy both requirements. The key is to use a null set supporting the singular part to recover each component directly from the original measure.

Throughout the main result, $(X,\mathcal{M},\mu)$ is a measure space with $\mu$ a positive measure; no finiteness or sigma-finiteness assumption on $\mu$ is needed for uniqueness. The measures $\nu,\alpha,\sigma$ are finite real signed measures on $(X,\mathcal{M})$. For a finite signed measure $\rho$, $|\rho|$ denotes its total variation measure. We say $\rho\ll\mu$ if $|\rho|(E)=0$ whenever $E\in\mathcal{M}$ and $\mu(E)=0$. We say $\rho\perp\mu$ if there is an $N\in\mathcal{M}$ such that $\mu(N)=0$ and $|\rho|(N^c)=0$. Thus singularity means concentration, up to total variation zero, on a measurable $\mu$-null set.

Theorem (Uniqueness of the Lebesgue Decomposition): Let $(X,\mathcal{M},\mu)$ be a measure space with $\mu$ positive, and let $\nu$ be a finite real signed measure. Suppose $$ \nu=\alpha_1+\sigma_1=\alpha_2+\sigma_2, $$ where $\alpha_i\ll\mu$ and $\sigma_i\perp\mu$ for $i=1,2$, with all four component measures finite signed measures. Then $\alpha_1=\alpha_2$ and $\sigma_1=\sigma_2$ as measures: their values agree on every set in $\mathcal{M}$.

This includes the usual decomposition of a finite positive measure: its absolutely continuous and singular components are positive measures, and hence are also finite signed measures. The theorem is a uniqueness statement conditional on a decomposition being given; it does not assert existence under additional hypotheses.

Recovering the Parts from a Null Set

For each singular component choose a measurable null set on which it is concentrated. Their union is still null and supports both singular components. On that union, the absolutely continuous components vanish; off it, the singular components vanish. This gives explicit recovery formulas.

Proof. Since $\sigma_1\perp\mu$, there is $N_1\in\mathcal{M}$ with $\mu(N_1)=0$ and $|\sigma_1|(N_1^c)=0$. Since $\sigma_2\perp\mu$, there is $N_2\in\mathcal{M}$ with $\mu(N_2)=0$ and $|\sigma_2|(N_2^c)=0$. Set $N=N_1\cup N_2$. Then $N\in\mathcal{M}$, $\mu(N)=0$, and $|\sigma_i|(N^c)=0$ for $i=1,2$, because $N^c\subseteq N_i^c$. Also, $|\alpha_i|(N)=0$, by $\alpha_i\ll\mu$.

Fix any $E\in\mathcal{M}$. On $E\setminus N=E\cap N^c$, both singular components have total variation zero, so the two decompositions give $\nu(E\setminus N)=\alpha_i(E\setminus N)$ for $i=1,2$. Consequently $\alpha_1(E\setminus N)=\alpha_2(E\setminus N)$. Since $|\alpha_i|(N)=0$, we have $\alpha_i(E)=\alpha_i(E\setminus N)$, so $\alpha_1(E)=\alpha_2(E)$. On $E\cap N$, both absolutely continuous components have total variation zero, so $\nu(E\cap N)=\sigma_i(E\cap N)$ for $i=1,2$. The singular components vanish in total variation on $N^c$, giving $\sigma_i(E)=\sigma_i(E\cap N)$. Therefore $\sigma_1(E)=\sigma_2(E)$. These equalities hold for every $E\in\mathcal{M}$, which proves equality of the measures. The argument also covers $N=\varnothing$: then the singular components vanish and both absolutely continuous components equal $\nu$. $\square$

Recovery formulas: If $\nu=\alpha+\sigma$, where $\alpha\ll\mu$ and $\sigma\perp\mu$, and $N\in\mathcal{M}$ satisfies $\mu(N)=0$ and $|\sigma|(N^c)=0$, then for every $E\in\mathcal{M}$, $$ \alpha(E)=\nu(E\setminus N),\qquad \sigma(E)=\nu(E\cap N). $$

Proof. Absolute continuity gives $|\alpha|(N)=0$, so $\alpha(E)=\alpha(E\setminus N)$. Singularity gives $|\sigma|(N^c)=0$, so $\sigma(E)=\sigma(E\cap N)$. On $E\setminus N$, the decomposition $\nu=\alpha+\sigma$ therefore reduces to $\nu(E\setminus N)=\alpha(E\setminus N)$. On $E\cap N$, it reduces to $\nu(E\cap N)=\sigma(E\cap N)$. Combining these equalities proves both formulas. $\square$

The recovery formulas also explain why the choice of supporting null set cannot affect the answer: any such choice recovers the same components, since the theorem proves equality on every measurable set. The decomposition does not require a unique null support; it requires unique measures as components.

Worked Example: Lebesgue Measure Plus a Point Mass

Let $X=\mathbb{R}$, $\mathcal{M}=\mathcal{L}$, and let $\mu=m$ be Lebesgue measure, which is complete and sigma-finite. Define the finite positive measure $\nu(E)=2m(E\cap[0,1])+3\delta_0(E)$ for $E\in\mathcal{L}$, where $\delta_0$ is the unit point mass at $0$. Its proposed components are $\alpha(E)=2m(E\cap[0,1])$ and $\sigma(E)=3\delta_0(E)$.

For every Lebesgue-null $E$, $\alpha(E)=2m(E\cap[0,1])=0$, so $\alpha\ll m$. The set $N=\{0\}$ satisfies $m(N)=0$, and $|\sigma|(N^c)=0$, so $\sigma\perp m$. To check the total masses, $m([0,1])=1$ and $\delta_0(\mathbb{R})=1$, giving $\alpha(\mathbb{R})=2$, $\sigma(\mathbb{R})=3$, and $\nu(\mathbb{R})=5$.

For $E=[0,1]$, $\alpha(E)=2m([0,1])=2$, $\sigma(E)=3\delta_0([0,1])=3$, and $\nu(E)=2+3=5$. For $E=(0,1]$, $\alpha(E)=2m((0,1])=2$, $\sigma(E)=3\delta_0((0,1])=0$, and $\nu(E)=2$. For $E=\{0\}$, $\alpha(E)=2m(\{0\})=0$, $\sigma(E)=3\delta_0(\{0\})=3$, and $\nu(E)=3$. The recovery formulas with $N=\{0\}$ yield the same components on all measurable sets. As an independent total-mass check, $\nu(\mathbb{R})=2m([0,1])+3\delta_0(\mathbb{R})=2+3=5=\alpha(\mathbb{R})+\sigma(\mathbb{R})$.

Worked Example: A Reference Measure with One Supported Atom

Let $X=\{0,1\}$, $\mathcal{M}=\mathcal{P}(X)$, and $\mu=\delta_0$, a finite positive measure. Define $\nu=7\delta_0+4\delta_1$. Then $\alpha=7\delta_0$ is absolutely continuous with respect to $\mu$, and $\sigma=4\delta_1$ is singular with respect to $\mu$.

Indeed, the only $\mu$-null subsets are $\varnothing$ and $\{1\}$, and $\alpha$ assigns zero to both. For singularity, take $N=\{1\}$: $\mu(N)=0$, and $|\sigma|(N^c)=|\sigma|(\{0\})=0$. For the four measurable sets, the component values are: on $\varnothing$, $(\alpha,\sigma,\nu)=(0,0,0)$; on $\{0\}$, they are $(7,0,7)$; on $\{1\}$, they are $(0,4,4)$; and on $X$, they are $(7,4,11)$. In particular, the recovery formulas give $\alpha(E)=\nu(E\setminus\{1\})$ and $\sigma(E)=\nu(E\cap\{1\})$. As a check, $\nu(X)=7+4=11=\alpha(X)+\sigma(X)$, and uniqueness guarantees these are the only components with the required properties.

Worked Example: Counting Measure Has No Nonempty Null Set

Let $X=\mathbb{N}=\{1,2,\ldots\}$, let $\mathcal{M}=\mathcal{P}(\mathbb{N})$, and let $\mu$ be counting measure. This measure is sigma-finite but not finite. Define the finite positive measure $\nu(E)=\sum_{n\in E}2^{-n}$. Its total mass is $\sum_{n=1}^{\infty}2^{-n}=1$.

Every nonempty $E\subseteq\mathbb{N}$ contains some $n$, so $\mu(E)\geq\mu(\{n\})=1$. Thus the only $\mu$-null set is $\varnothing$. Every $\nu$-null set is therefore also $\mu$-null, so $\nu\ll\mu$. The only measure singular with respect to $\mu$ is the zero measure: if a measure is concentrated on a $\mu$-null set, that set must be empty. Hence the decomposition is $\alpha=\nu$ and $\sigma=0$.

For a direct check, $\alpha(\{1,2\})=2^{-1}+2^{-2}=3/4$, while $\nu(\{1,2\})=3/4$ and $\sigma(\{1,2\})=0$. Also $\alpha(\mathbb{N})=\sum_{n=1}^{\infty}2^{-n}=1=\nu(\mathbb{N})$. The recovery formula with $N=\varnothing$ returns $\alpha(E)=\nu(E)$ and $\sigma(E)=0$ for every $E\subseteq\mathbb{N}$.

A Common Mistake: Confusing Support with Absolute Continuity

Absolute continuity does not mean that a measure is concentrated where the reference measure is concentrated. It means that every reference-null set receives zero measure. For example, on $(\mathbb{R},\mathcal{L},m)$, the set $\{0\}$ is null, but $\delta_0(\{0\})=1$. Thus $\delta_0\not\ll m$, even though it is concentrated entirely at one point. It is singular with respect to $m$, not absolutely continuous.

A wrong proposed split for $\nu=m|_{[0,1]}+\delta_0$ is to put the point mass into the absolutely continuous part, arguing that $0\in[0,1]$. This would assign $\delta_0$ to a measure claimed to be absolutely continuous with respect to $m$. But at the null set $E=\{0\}$, that proposed component gives mass $1$, contradicting the definition. The correction is to take $\alpha=m|_{[0,1]}$, $\sigma=\delta_0$, and $N=\{0\}$. The recovery formulas then identify the parts exactly.

Finally, uniqueness concerns the component measures, not pointwise values of a density. If the absolutely continuous part is expressed by a Radon–Nikodym density, that density is determined only up to equality $\mu$-a.e.; changing a density on a $\mu$-null set does not change its integral measure. This distinction is consistent with the theorem: the measures themselves agree on every set, while representatives of their densities need not agree everywhere.

Check Your Understanding

Use the null-set recovery argument to answer each question. Equalities of measures should be interpreted as equality on every measurable set.

Suppose $\nu=\alpha+\sigma$ on $(X,\mathcal{M},\mu)$, with $\alpha\ll\mu$ and $\sigma\perp\mu$. If $N$ is a null set supporting $\sigma$, write the recovery formulas for $\alpha(E)$ and $\sigma(E)$.
Why is the union of two null sets supporting two candidate singular components still sufficient for the uniqueness proof?
On $(\mathbb{R},\mathcal{L},m)$, can a nonzero multiple of $\delta_0$ be part of an absolutely continuous component? Check the definition on $\{0\}$.
For counting measure on $\mathbb{N}$, identify every null set and explain why every measure singular with respect to counting measure must be zero.
Why does uniqueness of an absolutely continuous measure not imply pointwise uniqueness of its Radon–Nikodym density?

Uniqueness of the Lebesgue Decomposition

What You'll Learn