Signed measures, densities

Published

18 October 2026

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This text is still being edited: it may contain typos, inaccuracies, and locally unpolished passages. If you find something, please let the author know (contact details on the course page).

1 Signed measures

1.1 Introduction

We want to understand what kinds of signed measures exist and how to decompose them into simple pieces. First we decompose into a positive/negative part (Jordan), then, relative to a reference measure, into a singular/absolutely continuous part (Lebesgue), and then for the absolutely continuous part we find a density (Radon-Nikodym).

Let $(X, \mathcal{F})$ be a measurable space.

Definition 1 We call a function $\nu : \mathcal{F} \to [-\infty , +\infty ]$ a real measure, a signed measure, or a charge if

$\nu (\varnothing ) = 0$;
$+\infty \not\in \operatorname {Im}\nu$ or $-\infty \not\in \operatorname {Im}\nu$. In other words: $\nu$ cannot take the values $+\infty$ and $-\infty$ simultaneously. That is, if $\nu$ takes the value $+\infty$ on some set, then it does not take the value $-\infty$ on any other set, and vice versa.
$\nu$ is countably additive: if $E_1, E_2, \ldots \in \mathcal{F}$ is a sequence of pairwise disjoint sets from the sigma-algebra, then

\[ \nu \left(\bigsqcup _{n=1}^{+\infty }E_n\right) = \sum _{n=1}^{+\infty }\nu \left(E_n\right) \]

Condition 2 is needed so that no indeterminate form $+\infty + (-\infty )$ arises in additivity.

The meaning of the word “measure” varies across the mathematical literature. As a rule, a “measure” is a countably additive function on a sigma-algebra $\mu : \mathcal{F} \to [0,+\infty ]$ such that $\mu (\varnothing ) = 0$ (this is how we defined “measure” at the very beginning). However, some authors use the word “measure” to refer to what we have just introduced under the name “real measure”; in that case they use the expression “positive measure” or “nonnegative measure” for what we defined as a “measure.” To remove this ambiguity, in this chapter we will call a “measure” a “(nonnegative) measure.” A (nonnegative) measure is a special case of a real measure (in our terminology).

As a rule, we will denote a (nonnegative) measure by the letter $\mu$, and a real measure by the letter $\nu$.

Some authors use the term “signed measure” instead of “real measure”; some authors do not allow a real measure to take the value $+\infty , -\infty$.

Let us examine countable additivity of a real measure in more detail.

Example 1

Let $E_1, E_2, \ldots \in \mathcal{F}$ be an arbitrary pairwise disjoint collection of sets. Investigate the convergence of the series $\sum_{n=1}^{+\infty }\nu (E_n)$.

Notice that now the terms of the series $\sum_{n=1}^{+\infty }\nu (E_n)$ may be negative.

Solution

Can this series diverge? (Diverge in the sense that it has several limit points, including infinities.) No! Indeed, this series must equal $\nu (E) \in [-\infty , +\infty ]$, where $E =\bigsqcup_{n=1}^{+\infty }E_n$. That is, the series $\sum_{n=1}^{+\infty }\nu (E_n)$ either converges to a finite number or to one of the infinities.
Suppose $\sum_{n=1}^{+\infty }\nu (E_n)$ converges to a finite number. Can it converge conditionally? Also no! Indeed, if $p: \mathbb {N} \to \mathbb {N}$ is some permutation of the natural numbers, then $\bigsqcup_{n=1}^{+\infty }E_n = \bigsqcup_{n=1}^{+\infty }E_{p_n}$. That is, it does not matter in what order we take the union of the sets. By the definition of countable additivity we obtain

\[ \sum _{n=1}^{+\infty } \nu \left(E_{n}\right)=\nu \left(\bigsqcup _{n=1}^{+\infty }E_n \right) = \nu \left( \bigsqcup _{n=1}^{+\infty }E_{p_n}\right) = \sum _{n=1}^{+\infty } \nu \left(E_{p_n}\right) \]

It follows that the sum of the series $\sum_{n=1}^{+\infty } a_n$ (where $a_n = \nu \left(E_{p_n}\right)$) does not depend on the permutation of its terms. By the Riemann series theorem, a series of real numbers converges (to a finite number) with its sum independent of the permutation of its terms (such convergence is called unconditional) if and only if it converges absolutely, i.e. the series of its absolute values converges.

Let us state the resulting lemma.

Lemma 1 Let $\nu$ be a real measure on $(X,\mathcal{F})$, and let $E_1, E_2, \ldots \in \mathcal{F}$ be a pairwise disjoint collection of sets. Then the series $\sum_{n=1}^{+\infty } \nu (E_n)$ converges unconditionally (to a finite number or to one of the infinities).

Moreover, if it converges to a finite number, then it converges absolutely.

How can real measures be obtained? Via the Lebesgue integral with respect to some (nonnegative) measure.

Lemma 2 Let $\mu$ be a (nonnegative) measure on $(X,\mathcal{F})$. Let $h: X \to \overline{\mathbb {R}}$ be a measurable function. Denote \[ \nu _h(E) := \int _{E} h \; d\mu \] (if the integral exists).

If $h$ is nonnegative (i.e. $\operatorname {Im}h \subset [0, +\infty ]$), then $\nu_h$ is a (nonnegative) measure;
If at least one of the functions $h^+$, $h^-$ is integrable, then $\nu_h$ is a real measure.

For $\nu_h$ one also uses the notation $\nu_h = h \; d\mu$, to emphasize the role of $\mu$.

In what cases is $\nu = \nu_h$ for some function $h$? See the Radon-Nikodym theorem below.

1.2 The Hahn theorem and the Jordan theorem: decomposing a real measure into positive and negative parts

Theorem 1 (Hahn decomposition, without proof) Let $\nu$ be a real measure on $(X, \mathcal{F})$. Then there exists a partition of $X$ into $X^+, X^- \in \mathcal{F}$ (i.e. $X = X^+ \sqcup X^-$) such that

$\nu (E) \geq 0$ for all $E \in \mathcal{F}$, $E \subset X^{+}$
$\nu (E) \leq 0$ for all $E \in \mathcal{F}$, $E \subset X^{-}$

Real or (nonnegative) measures $\nu_1$, $\nu_2$ are called singular if they “live” on disjoint sets, i.e. there exists a partition of $X$ into $X_1, X_2 \in \mathcal{F}$ ($X = X_1 \sqcup X_2$) such that

\[ \nu _1 (E) = \nu _1\left(E \cap X_1\right), \quad \nu _2(E) = \nu _{2}\left(E \cap X_2\right) \qquad \text{for all } E\in \mathcal{F} \]

Example 2

Show that $\lambda$ and $\#_{\mathbb {Z}}$ (counting measure on $\mathbb {Z}$) are singular (nonnegative) measures on $(\mathbb {R}, \mathscr {B}\left(\mathbb {R}\right))$.

Solution

Let $X_1 = \mathbb {R} \setminus \left\{ \mathbb {Z}\right\}$, $X_2 = \mathbb {Z}$.

Note that if in the Hahn theorem we define

\[ \nu ^+(E) := \nu (E \cap X^+), \qquad \nu ^{-}(E) := -\nu (E \cap X^-), \]

then $\nu^+$ and $\nu^-$ are exactly the singular (nonnegative) measures we wanted. Moreover $\nu = \nu^+ - \nu^-$. We have essentially proved

Theorem 2 (Jordan decomposition) Every real measure is the difference of two (nonnegative) singular measures, at least one of which is finite. Namely, if $\nu$ is a real measure on $(X, \mathcal{F})$, then there exist (nonnegative) singular measures $\nu^+$ and $\nu^-$ such that $\nu = \nu^+ - \nu^-$.

$\nu$ may take the value $+\infty$ or $-\infty$, but not both. Hence, among the (nonnegative) measures $\nu^+$, $\nu^-$ at least one is finite. Moreover, they are both finite (do not take the value $+\infty$) if and only if $\nu$ is finite ($+\infty , -\infty \not\in \operatorname {Im}\nu$).

The (nonnegative) measure $\left|\nu \right| := \nu^+ + \nu^{-}$ is called the total variation measure of $\nu$.

If $d\nu = h \; d\mu$ for some measurable $h$, then it is not hard to show that

\[ d\nu ^+ = h^+ \; d\mu , \qquad d\nu ^- = h^- \; d\mu , \qquad d\left|\nu \right| = \left|h\right| \; d\mu \]

That is,

\[ \nu ^+(E) = \int _{E} h^{+} \; d\mu , \qquad \nu ^-(E) = \int _{E} h^{-} \; d\mu , \qquad \left|\nu \right|(E) = \int _{E} \left|h\right| \; d\mu \]

for all $E \in \mathcal{F}$.

Problem 1

On the measurable space $(\mathbb {R}, \mathscr {B}\left(\mathbb {R}\right))$ let

$d\nu_1 = f \; d\lambda$, where $f_1(x) = x^2 - 4$
$\nu_2 = \sum_{k=-3}^{3}(-1)^k\cdot k \cdot \delta_{k}$

Let $\nu = \nu_1 + \nu_2$.

Find the Jordan decomposition $\nu = \nu^+ - \nu^-$ of the real measure $\nu$.
Find the integral $\int_{[-2,0]} f \; d\left|\nu \right|$, where $f(x) = x^2$.

Problem 2

On the measurable space $(\mathbb {R}, \mathscr {B}\left(\mathbb {R}\right))$ let

$d\nu_1 = h_1\; d\lambda$, where $h_1(x) = x^2 - 4$
$d\nu_2 = h_2 \; d\lambda$, where $h_2(x) = (x-2)^2 - 5$

Let $\nu = \nu_1 + \nu_2$.

Find the Jordan decomposition $\nu = \nu^+ - \nu^-$ of the real measure $\nu$.
Find the integral $\int_{[-2,0]} f \; d\left|\nu \right|$, where $f(x) = x^2$.

2 Measures obtained via the integral of another (nonnegative) measure

2.1 The Radon-Nikodym theorem: characterizing measures obtained via the integral of another measure

$\mu$ will everywhere be a (nonnegative) measure, while $\nu$ will first be (nonnegative), then a real measure. Both will be $\sigma$-finite. Without the sigma-finiteness condition, various pathological counterexamples to the theorem below arise.

2.1.1 The case of a (nonnegative) measure $\nu$

Recall that if $(X,\mathcal{F}, \mu )$ is a space with a (nonnegative) measure and $h: X \to [0,+\infty ]$ is a nonnegative function, then

\[ \nu _h(E) := \int _E h \; d\mu \]

is also a (nonnegative) measure. Moreover, if $\mu (E) = 0$, then $\nu_h(E) = 0$. This property is called absolute continuity. Namely, if $\mu$, $\nu$ are two (nonnegative) measures, one says that $\nu$ is absolutely continuous with respect to $\mu$, or $\nu$ is dominated by $\mu$, and writes $\nu \ll \mu$, if $\nu (E) = 0$ for all sets of $\mu$-measure zero.

Let us connect this definition with the more familiar $\varepsilon$-$\delta$ construction.

Lemma 3 (An equivalent condition for absolute continuity of a finite (nonnegative) measure) Let $\nu$ be a finite (nonnegative) measure, and $\mu$ an arbitrary (nonnegative) measure. The following conditions are equivalent:

$\nu \ll \mu$
For every $\varepsilon > 0$ there exists $\delta > 0$ such that $\nu (E) < \varepsilon$ for all measurable $E$ such that $\mu (E) < \delta$.

It turns out that in the sigma-finite case, absolute continuity is equivalent to the measure being representable via an integral!

Theorem 3 (Radon-Nikodym) Let $\mu$, $\nu$ be (nonnegative) $\sigma$-finite measures on $(X,\mathcal F)$. The following conditions are equivalent:

$\nu \ll \mu$;
$d\nu = h \; d\mu$ for some nonnegative measurable $h: X \to [0,+\infty ]$

Moreover, $h$ is unique up to agreement $\mu$-a.e. That is, if $g$ is another function such that $\nu (E) = \int_E g \; d\mu$, then $h = g$ $\mu$-a.e.

Proof

$\Rightarrow$ (i) is obvious: if $d\nu = h\, d\mu$ and $\mu (E)=0$, then $\nu (E) = \int_E h\, d\mu = 0$.

Let us prove (i) $\Rightarrow$ (ii). The idea (due to J. von Neumann) is to apply the Riesz representation theorem for the Hilbert space $L^2(\mu +\nu )$ (the Riesz representation theorem, the chapter on the Hilbert space $L^2(\mu )$); this is exactly the point in the course where the study of Hilbert spaces pays off.

Step 1 (existence, finite measures). Suppose first that $\mu (X), \nu (X) < \infty$. Then $\mu +\nu$ is a finite measure, and $L^2(\mu +\nu )$ is a Hilbert space with inner product $\left\langle f, g \right\rangle = \int fg \; d(\mu +\nu )$ (the Riesz-Fischer theorem, the seminar on $L^p$-spaces). Define the linear functional

\[ \varphi (f) := \int f \; d\nu , \qquad f \in L^2(\mu +\nu ). \]

It is well defined: since $\nu \ll \mu$, and hence also $\nu \ll \mu +\nu$, functions that agree $(\mu +\nu )$-a.e. also agree $\nu$-a.e., so the value $\varphi (f)$ does not depend on the choice of representative of the equivalence class. It is bounded: applying the Cauchy-Bunyakovsky-Schwarz inequality (theorem the Cauchy-Bunyakovsky-Schwarz inequality) to the functions $\left|f\right|$ and $\mathbb 1$ (the latter lies in $L^2(\mu +\nu )$, since the measure $\mu +\nu$ is finite),

\[ \left|\varphi (f)\right| \; \leq \; \int \left|f\right| \; d\nu \; \leq \; \int \left|f\right| \; d(\mu +\nu ) \; \leq \; \left(\mu (X)+\nu (X)\right)^{1/2} \cdot \left\| f\right\| _{2} \; < \; \infty \]

(the step $\int \left|f\right| \, d\nu \leq \int \left|f\right| \, d(\mu +\nu )$ holds because $\mu \geq 0$). Hence $\varphi$ is a bounded linear functional on $L^2(\mu +\nu )$. By the Riesz representation theorem there is $g \in L^2(\mu +\nu )$ such that

\[ \int f \; d\nu \; = \; \int fg \; d(\mu +\nu ), \qquad \forall f \in L^2(\mu +\nu ), \]

that is,

\[ \int f(1-g) \; d\nu \; = \; \int fg \; d\mu , \qquad \forall f \in L^2(\mu +\nu ). \tag {$*$} \]

Substitute into ($*$) $f = \; \mathbb {1}_{g \geq 1}$: on the left $\int_{\left\{ g\geq 1\right\} }(1-g)\, d\nu \leq 0$ (since $1-g\leq 0$ on $\left\{ g\geq 1\right\}$, $\nu \geq 0$), on the right $\int_{\left\{ g\geq 1\right\} } g\, d\mu \geq 0$ (since $g\geq 1>0$ there, $\mu \geq 0$). Both sides are equal and are sandwiched between $\leq 0$ and $\geq 0$, so both are zero; from the right-hand side being zero and $g\geq 1$ on $\left\{ g\geq 1\right\}$, it follows that $\mu (\left\{ g\geq 1\right\} )=0$. Similarly, substituting $f=\; \mathbb {1}_{g<0}$: on the left $\int_{\left\{ g<0\right\} }(1-g)\, d\nu \geq 0$ (since $1-g>1>0$ there), on the right $\int_{\left\{ g<0\right\} }g\, d\mu \leq 0$ (since $g<0$ there); the same reasoning gives $\mu (\left\{ g<0\right\} )=0$.

Since $\nu \ll \mu$, from $\mu (\left\{ g\geq 1\right\} ) = \mu (\left\{ g<0\right\} ) = 0$ it also follows that $\nu (\left\{ g\geq 1\right\} ) = \nu (\left\{ g<0\right\} ) = 0$. Redefine $g$, setting it equal to $\tfrac 12$ on the set $\left\{ g\geq 1\right\} \cup \left\{ g<0\right\}$ — this set has zero $(\mu +\nu )$-measure, so the identity ($*$) does not change — and now $0\leq g < 1$ everywhere on $X$. Set

\[ h := \frac{g}{1-g} \; \geq \; 0 \]

(the denominator is strictly positive everywhere), and let us prove that $d\nu =h\, d\mu$. Fix $E\in \mathcal F$ and for $k\in \mathbb {N}$ set

\[ f_k(x) := \begin{cases} \dfrac {\; \mathbb {1}_{E}(x)}{1-g(x)}, & \text{if } \dfrac {\; \mathbb {1}_{E}(x)}{1-g(x)} \leq k, \\ 0, & \text{otherwise.} \end{cases} \]

The function $f_k$ is bounded by $k$, hence $f_k \in L^2(\mu +\nu )$ (the measure $\mu +\nu$ is finite). Substituting $f_k$ into ($*$): $\int f_k(1-g)\, d\nu = \int f_k g \, d\mu$. For fixed $x$ the sequence $f_k(x)$ is nondecreasing in $k$ (from $0$ up to the value $\tfrac {\; \mathbb {1}_{E}(x)}{1-g(x)}$, at which it stays for all further $k$); hence $f_k(1-g) \uparrow \; \mathbb {1}_{E}$ and $f_kg \uparrow \; \mathbb {1}_{E} \cdot h$ pointwise and monotonically. By Levi’s monotone convergence theorem, passing to the limit as $k\to \infty$:

\[ \nu (E) \; = \; \int _E 1 \; d\nu \; = \; \int _E h \; d\mu . \]

Since $E\in \mathcal F$ was arbitrary, $d\nu = h\, d\mu$. This completes Step 1.

Uniqueness. Let $h,\widetilde h : X \to [0,+\infty ]$ both be measurable and $\nu (E) = \int_E h\, d\mu = \int_E \widetilde h\, d\mu$ for all $E\in \mathcal F$. Since $\mu$ is $\sigma$-finite, there exist $X_1\subseteq X_2\subseteq \cdots \in \mathcal F$, $\bigcup_k X_k = X$, $\mu (X_k)<\infty$. Fix $k$ and set $E := \left\{ h>\widetilde h\right\} \cap X_k$; then $\int_E (h-\widetilde h)\, d\mu = \nu (E)-\nu (E) = 0$, and the integrand is strictly positive on $E$ with $\mu (E)\leq \mu (X_k)<\infty$ — a standard fact about the Lebesgue integral gives $\mu (E)=0$. Similarly $\mu (\left\{ h<\widetilde h\right\} \cap X_k)=0$. Letting $k\to \infty$ (continuity of the measure from below): $\mu (\left\{ h>\widetilde h\right\} ) = \mu (\left\{ h<\widetilde h\right\} ) = 0$, i.e. $h=\widetilde h$ $\mu$-a.e.

Step 2 (existence, $\sigma$-finite measures). Now suppose $\mu ,\nu$ are only $\sigma$-finite. Since both are $\sigma$-finite, one can choose (by intersecting exhausting sequences for $\mu$ and for $\nu$) a single increasing sequence $X_1\subseteq X_2\subseteq \cdots \in \mathcal F$ such that $\bigcup_k X_k = X$ and $\mu (X_k), \nu (X_k) < \infty$ for each $k$. Let $\mathcal F_k := \left\{ E \cap X_k \; : \; E \in \mathcal F\right\}$, and let $\mu_k, \nu_k$ be the restrictions of $\mu , \nu$ to $(X_k, \mathcal F_k)$; these are finite measures, and $\nu_k \ll \mu_k$ (immediately from $\nu \ll \mu$). By Step 1 there is a nonnegative measurable $h_k : X_k \to [0,+\infty )$ such that $\nu_k(E) = \int_E h_k \, d\mu_k$ for all $E \in \mathcal F_k$. By the already-established uniqueness, for $j<k$ the functions $h_j$ and $h_k$ both represent $\nu_j = \nu_k|_{\mathcal F_j}$ with respect to $\mu_j = \mu_k|_{\mathcal F_j}$, so $h_j = h_k$ $\mu$-a.e. on $X_j$.

Define $h: X \to [0,+\infty )$ as follows: $h(x) := h_k(x)$ for $x \in X_k$ (this definition is consistent $\mu$-a.e. by what was just said; on the remaining set of $\mu$-measure $0$, define $h$ to be zero). For an arbitrary $E \in \mathcal F$, by continuity of the measure from below ($E = \bigcup_k (E\cap X_k)$, the sets increasing) and Levi’s monotone convergence theorem,

\[ \nu (E) \; = \; \lim _{k\to \infty } \nu (E \cap X_k) \; = \; \lim _{k\to \infty }\int _{E\cap X_k} h_k \; d\mu \; = \; \lim _{k\to \infty }\int _E h \cdot \; \mathbb {1}_{X_k} \; d\mu \; = \; \int _E h \; d\mu . \]

Hence $d\nu = h\, d\mu$. This completes the proof of existence, and hence of the whole theorem (uniqueness in the general $\sigma$-finite case was already established above).

Remark

2.2 a preview: change of density on path space — the Cameron-Martin and Girsanov theorems

The Radon-Nikodym theorem is proved for an arbitrary measurable space $(X,\mathcal F)$. As $X$ one may also take the space of continuous functions $C[0,1]$ with the uniform metric (see the chapter on the Lebesgue spaces $L^p$, where it was discussed that $C[0,1]$ is the natural phase space for the paths of random processes) — and then the Radon-Nikodym theorem becomes a substantive tool of the theory of random processes, not just of analysis on $\mathbb {R}^n$.

Let $\mathbb {P}$ be the Wiener measure on $C[0,1]$, i.e. the distribution of the path of standard Brownian motion $(W_t)_{t\in [0,1]}$, $W_0=0$. Fix a constant $c\in \mathbb {R}$ and consider the “drifted” Brownian motion $W_t + ct$; let $\mathbb {Q}$ be its distribution (also a measure on $C[0,1]$). The simplest special case of Girsanov’s theorem is the Cameron-Martin theorem: $\mathbb {Q} \ll \mathbb {P}$, and moreover

\[ \frac{d\mathbb {Q}}{d\mathbb {P}}(\omega ) = \exp \left(c\, \omega (1) - \frac{c^2}{2}\right), \qquad \omega \in C[0,1]. \]

In other words, to compute the probability of any event concerning the drifted Brownian motion, it suffices to take the same event for ordinary Brownian motion and multiply the integrand by the explicit, computable density $\exp (c\, \omega (1)-c^2/2)$, which depends only on the value of the path $\omega (1)$ at the final time. The formula can plausibly be checked on finite-dimensional distributions: the vector $(W_{t_1},\ldots ,W_{t_n})$ has a multivariate Gaussian density, and by direct computation (shifting the mean in the Gaussian density) the density of the shifted vector relative to the ordinary one indeed turns out to be the exponential of a linear function of the values.

The general Girsanov theorem replaces the constant drift $ct$ by the drift $\int_0^t \theta_s\, ds$ for an arbitrary (adapted) random process $\theta$, and the density formula turns into the exponential martingale $\exp \left(\int_0^1 \theta_s\, dW_s - \frac12\int_0^1\theta_s^2\, ds\right)$ instead of the simple $c\, \omega (1)-c^2/2$. To make sense of this integral with respect to Brownian motion ($dW_s$, rather than $ds$) one needs the Itô stochastic integral — the subject of a separate course on random processes. But the principle behind this preview is exactly the one we have already proved: the Radon-Nikodym theorem, applied not to $\mathbb {R}^n$ but to path space.

Notation: $h = \frac{d\nu }{d\mu }$. $h$ is called the Radon-Nikodym derivative.

Lemma 4 (Properties of the Radon-Nikodym derivative) Let $\mu$ be a (nonnegative) $\sigma$-finite measure, and let $\nu_1$, $\nu_2$ also be (nonnegative) $\sigma$-finite measures. Then

if $\nu_1 \ll \mu$ and $\nu_2 \ll \mu$, then

\[ \frac{d(\nu _1 + \nu _2)}{d\mu } = \frac{d\nu _1}{d\mu } + \frac{d\nu _2}{d\mu } \qquad \mu \text{-a.e.} \]
if $\nu_1 \ll \nu_2 \ll \mu$, then

\[ \frac{d\nu _1}{d\mu } = \frac{d\nu _1}{d\nu _2} \cdot \frac{d\nu _2}{d\mu } \qquad \mu \text{-a.e.} \]
if $\nu_1 \ll \nu_2$ and $\nu_2 \ll \nu_1$, then

\[ \frac{d\nu _2}{d\nu _1} = \left(\frac{d\nu _1}{d\nu _2}\right)^{-1} \qquad \nu _1\text{-a.e.} \]

Problem 3

Consider on $(\mathbb {R}, \mathscr {B}\left(\mathbb {R}\right))$ the following (nonnegative) measures:

$\lambda$ (Lebesgue measure on $\mathbb {R}$);
$\frac{1}{3} \cdot \lambda_{[-1,1]}$ (Lebesgue measure on $[-1,1]$ with a coefficient);
$f \; d\lambda$, where $f(x) = x^{2}I_{[0,1]}(x)$;
$f \; d\lambda + 3\delta_{7}$, where $f(x) = x^{2}I_{[0,1]}(x)$;
$f \; d\lambda$, where $f(x) = e^{-3\left|x\right|}I_{[-1,1]}(x)$;
$f \; d\lambda$, where $f(x) = \frac{1}{x} \cdot I_{(0,1]}(x)$;
$\#_{\mathbb {N}}$;
$\#_{\mathbb {Z}}$;
$\#_{\mathbb {Q}}$;
$\delta_0$;
$7\cdot \delta_{-3} + \frac{1}{3}\delta_{2}$; Check that they are sigma-finite. Then go through all possible pairs $\nu , \mu$ among them and check whether $\nu \perp \mu$ or $\nu \ll \mu$ holds. If the latter holds, find $\frac{d\nu }{d\mu }$.

The same question makes sense not only on $\mathbb {R}$: any measurable (in particular, continuous) injection $\varphi : \mathbb {R}^n \to X$ from a finite-dimensional space into an arbitrary measurable space $(X, \mathcal F)$ allows one to transport a measure from $\mathbb {R}^n$ to $X$ — in exactly the same way that the distribution of a random variable transports the probability measure from $\Omega$ to $\mathbb {R}$. The following problem carries this out for $X = C[0,1]$ (with the Borel sigma-algebra with respect to the uniform metric, the chapter on the Lebesgue spaces $L^p$) — an infinite-dimensional space — but all the computations in it are arranged so as to reduce to the already familiar one- and two-dimensional Lebesgue measure.

Problem 4

Let $X = C[0,1]$. Denote by $\mathbf{0} \in X$ the identically zero function, and for $n \in \mathbb {N}$ let $f_n(x) := x^n$ (so that $f_1(x)=x$). Let $\varphi : \mathbb {R} \to X$, $\varphi (a) :=$ (the constant function $\equiv a$), and $\psi : \mathbb {R}^2 \to X$, $\psi (a,b) := \left(x \mapsto a + bx\right)$ (both are continuous, and hence measurable, injections). Consider on $(X, \mathcal B)$ the following (nonnegative) measures:

$\delta_{\mathbf0}$;
$\mu_1(A) := \lambda \left(\varphi^{-1}(A) \cap [0,1]\right)$ (transport of Lebesgue measure onto the line of constants);
$\mu_2(A) := \lambda_2\left(\psi^{-1}(A) \cap [0,1]^2\right)$ (transport of two-dimensional Lebesgue measure onto the plane of affine functions);
$\mu_3(A) := \int_{\psi^{-1}(A) \cap [0,1]^2} (a+b) \; d\lambda_2(a,b)$ (the same plane, but with density $(a+b)$);
$\mu_5(A) := \lambda_2\left(\psi^{-1}(A) \cap [0,\tfrac 12]^2\right)$ (the same plane, but with a truncated parameter domain);
$\mu_4 := \sum_{n=1}^\infty \frac{1}{2^n}\delta_{f_n}$ (a weighted sum of point masses at monomials);
$\#$ — counting measure on the countable set $\left\{ f_n \; : \; n \in \mathbb {N}\right\} \subset X$;
$\delta_g$, where $g(x) := \sin \left(\frac{\pi x}{2}\right)$. Go through all pairs $\nu , \mu$ from this list and check whether $\nu \perp \mu$ or $\nu \ll \mu$ holds; if the latter holds, find $\frac{d\nu }{d\mu }$ — if possible not in terms of the parameters $(a,b)$, but intrinsically, via the function $f \in X$ itself (say, via $f(0)$, $f(1)$, or the index $n$, if $f=f_n$).

To check that $\mu_1, \mu_2, \mu_3, \mu_5$ are indeed well-defined measures on $\mathcal B$, it suffices to note that countable additivity transfers one-to-one through an injective preimage from $\lambda$ (resp. $\lambda_2$) — this is the same fact you used when checking that the distribution of a random variable is a measure. For comparing measures “along the same $\psi$” (the pairs $\mu_2,\mu_3,\mu_5$), the comparison reduces to comparing densities/regions on $\mathbb {R}^2$ — just as in ordinary $\mathbb {R}^n$. A single point $\left\{ f\right\}$ always has zero $\mu_1,\mu_2,\mu_3,\mu_5$-measure (the preimage of a point under an injection is a point, which has $\lambda$- or $\lambda_2$-measure zero) — use this for all pairs involving a point mass. Also note that $a = f(0)$, $b = f(1)-f(0)$ for $f = \psi (a,b) = a+bx$ — this allows any answer using $(a,b)$ to be rewritten in terms of the values of $f$ at the points $0$ and $1$.

2.2.1 The case of a real measure $\nu$

Recall that if $(X,\mathcal{F}, \mu )$ is a space with a (nonnegative) measure and $h: X \to \mathbb {R}$ is a function such that at least one of $h^+$ and $h^-$ is integrable, then

\[ \nu _h(E) := \int _E h \; d\mu \]

is a real measure. Moreover, if $\mu (E) = 0$, then $\left|\nu_h\right|(E) = \int_E \left|h\right| \; d\mu = 0$.

If $\mu$ is a nonnegative measure and $\nu$ is a real measure, one says that $\nu$ is absolutely continuous with respect to $\mu$, or $\nu$ is dominated by $\mu$, and writes $\nu \ll \mu$, if $\left|\nu \right| \ll \mu$ in the sense of the definition above, i.e. if $\left|\nu \right|(E) = 0$ for all sets of $\mu$-measure zero. It is not hard to show that this is equivalent to $\nu^+ \ll \mu$ and $\nu^- \ll \mu$.

In summary, briefly

\[ \nu \ll \mu \quad \stackrel{\text{def}}{\iff } \quad \left|\nu \right| \ll \mu \quad \iff \quad \begin{cases} \nu ^+ \ll \mu \\ \nu ^- \ll \mu \end{cases} \]

It turns out that, again in the sigma-finite case, absolute continuity is equivalent to the measure being representable via an integral! Indeed, if $\nu \ll \mu$, then $\nu^+ \ll \mu$ and $\nu^- \ll \mu$, and hence by the Radon-Nikodym theorem for the previous case there are densities $h^+$, $h^-$.

Theorem 4 (Radon-Nikodym) Let $\mu$ be a $\sigma$-finite (nonnegative) measure, $\nu$ a real $\sigma$-finite measure, $\nu \ll \mu$. Then there exists a measurable function $h: X \to \overline{\mathbb {R}}$ such that \[ \nu (E) = \int _E h \; d\mu , \qquad \forall E \in \mathcal{F} \]

2.3 The Lebesgue theorem: decomposing a measure into an absolutely continuous and a singular part

$\mu$ will everywhere be a (nonnegative) measure, while $\nu$ will first be (nonnegative), then a real measure. Both will be $\sigma$-finite.

2.3.1 The case of a (nonnegative) measure $\nu$

Theorem 5 (Lebesgue) Let $\mu$ be a (nonnegative) $\sigma$-finite measure on $(X, \mathcal{F})$. Then for any (nonnegative) $\sigma$-finite measure $\nu$ on $(X, \mathcal{F})$ there exist unique (nonnegative) measures $\nu_{a}$, $\nu_{s}$ such that $\nu = \nu_{a} + \nu_{s}$ and \[ \nu _a \ll \mu , \qquad \nu _s \perp \mu \]

2.3.2 The case of a real measure $\nu$

Now let $\nu$ be a real $\sigma$-finite measure. Let $\nu = \nu^+ - \nu^-$ be its Jordan decomposition, consisting of $\sigma$-finite (nonnegative) measures. Each of $\nu^+$ and $\nu^-$ can be decomposed, by the Lebesgue theorem above, into an absolutely continuous and a singular component with respect to $\mu$:

\[ \nu ^+ = \nu ^+_a + \nu ^+_{s}, \qquad \nu ^- = \nu ^-_a + \nu ^-_{s} \]

Let

\[ \nu _a := \nu ^+_a - \nu ^-_a, \qquad \nu _{s} := \nu ^+_{s} - \nu ^-_{s} \]

What can be said about the real measures $\nu_a$, $\nu_{s}$?

$\left|\nu_a\right| = \nu^+_a + \nu^-_a \ll \mu$. That is, the real measure $\nu_a$ is absolutely continuous with respect to the (nonnegative) measure $\mu$.
$\nu^+_{s}$, $\nu^-_{s}$ are (nonnegative) measures, singular with respect to $\mu$. That is, there exist $A_1, B_1, A_2, B_2\in \mathcal{F}$: $X = A_1 \sqcup B_1 = A_2 \sqcup B_2$, with $\nu^+_{s}$ “living” on $A_1$ (in the sense that $\nu^+_{s}(E) = \nu^+_{s}(E \cap A_1)$ for any $E \in \mathcal{F}$), $\nu^-_{s}$ “living” on $A_2$, $\mu$ “living” on $B_1$ and $B_2$. Hence $\mu$ “lives” on $B_1 \cap B_2$, and $\nu_{s} = \nu^+_{s} - \nu^-_{s}$ “lives” on $A_1 \cup A_2$. Moreover

\[ X = \left(A_1 \cup A_2\right) \sqcup \left(B_1 \cap B_2\right) \]

In summary, $\nu_{s}$ and $\mu$ are singular.

Theorem 6 (Lebesgue) Let $\mu$ be a (nonnegative) $\sigma$-finite measure on $(X, \mathcal{F})$. Then for any real $\sigma$-finite measure $\nu$ on $(X, \mathcal{F})$ there exist unique real measures $\nu_{a}$, $\nu_{s}$ such that $\nu = \nu_{a} + \nu_{s}$ and \[ \nu _a \ll \mu , \qquad \nu _s \perp \mu \]

Remark

Everything we have discussed in this chapter is abstract and applies to arbitrary measurable spaces. The application of this theory to the concrete case $X=\mathbb {R}$ with Lebesgue measure — the relationship between absolute continuity of measures and absolute continuity of functions, and differentiation of a distribution’s density — is deferred to a separate chapter on differentiation of the Lebesgue integral, since that is also where all the analytic machinery naturally belongs (the Hardy-Littlewood maximal function, the Lebesgue differentiation theorem) needed for an honest proof.

Signed measures, densities

1 Signed measures

1.1 Introduction

1.2 The Hahn theorem and the Jordan theorem: decomposing a real measure into positive and negative parts

2 Measures obtained via the integral of another (nonnegative) measure

2.1 The Radon-Nikodym theorem: characterizing measures obtained via the integral of another measure

2.1.1 The case of a (nonnegative) measure \(\nu\)

2.2 a preview: change of density on path space — the Cameron-Martin and Girsanov theorems

2.2.1 The case of a real measure \(\nu\)

2.3 The Lebesgue theorem: decomposing a measure into an absolutely continuous and a singular part

2.3.1 The case of a (nonnegative) measure \(\nu\)

2.3.2 The case of a real measure \(\nu\)