4.2 Absolute Continuity and the Radon-Nikodym Theorem

In the previous section, we explored the internal structure of a single signed measure (the Hahn-Jordan decomposition). In this section, we shift our perspective to the relationship between two different measures on the same measurable space. If "mutually singular" characterizes the extreme case where two measures are completely unrelated, then "absolute continuity" describes the other extreme: one measure is entirely governed by another.

The pinnacle of exploring this relationship is a core pillar of modern analysis and probability theory—the Lebesgue-Radon-Nikodym Theorem.

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1. Absolute Continuity of Measures

In calculus, absolutely continuous functions possess favorable properties. In measure theory, absolute continuity reflects the "heritability of null sets."

Definition 4.2.1 (Absolute Continuity)

Let \(\mu\) be a non-negative measure on a measurable space \((X, \mathcal{M})\), and \(\nu\) be a signed measure (or complex measure) on the same space.

If for any measurable set \(E \in \mathcal{M}\) satisfying \(\mu(E) = 0\), we have:

\[ \nu(E) = 0 \]

then \(\nu\) is said to be absolutely continuous with respect to \(\mu\), denoted as \(\nu \ll \mu\).

• Note: Since the value of the total variation measure \(|\nu|\) on \(E\) is determined by the values of \(\nu\) on subsets of \(E\), \(\nu \ll \mu\) if and only if \(|\nu| \ll \mu\).

Absolute continuity has an extremely intuitive \(\epsilon-\delta\) characterization, which explains the origin of the term "continuity": as long as the \(\mu\)-measure of a set is sufficiently small, its \(\nu\)-measure can be made arbitrarily small.

Theorem 4.2.1 (\(\epsilon-\delta\) Characterization of Absolute Continuity)

Let \(\nu\) be a finite signed measure (i.e., \(|\nu|(X) < \infty\)). Then \(\nu \ll \mu\) if and only if: for any \(\epsilon > 0\), there exists \(\delta > 0\) such that for any measurable set \(E \in \mathcal{M}\) satisfying \(\mu(E) < \delta\), we have:

\[ |\nu(E)| < \epsilon \]

Proof of Theorem 4.2.1

Sufficiency (\(\Leftarrow\)): If \(\mu(E) = 0\), then for any \(\delta > 0\), the condition \(\mu(E) < \delta\) is satisfied. From the condition, it follows that for any \(\epsilon > 0\), \(|\nu(E)| < \epsilon\); therefore, we must have \(\nu(E) = 0\), i.e., \(\nu \ll \mu\).

Necessity (\(\Rightarrow\)): We use proof by contradiction. Suppose the conclusion does not hold; then there exists some \(\epsilon > 0\) such that for every \(n \in \mathbb{N}\) (taking \(\delta = 1/2^n\)), there exists a corresponding set \(E_n \in \mathcal{M}\) satisfying:

\[ \mu(E_n) < \frac{1}{2^n} \quad \text{but} \quad |\nu(E_n)| \ge \epsilon \]

Define the "limit superior" of the sequence of sets:

\[ F_k = \bigcup_{n=k}^\infty E_n, \quad F = \bigcap_{k=1}^\infty F_k \]

Due to the subadditivity of the measure \(\mu\):

\[ \mu(F_k) \le \sum_{n=k}^\infty \mu(E_n) < \sum_{n=k}^\infty \frac{1}{2^n} = \frac{1}{2^{k-1}} \]

Clearly, as \(k \to \infty\), \(\mu(F_k) \to 0\). By continuity from above, we have \(\mu(F) = 0\).

Since it is known that \(\nu \ll \mu\), we must have \(|\nu|(F) = 0\).

However, since \(\nu\) is a finite measure, utilizing continuity from above:

\[ |\nu|(F) = \lim_{k \to \infty} |\nu|(F_k) \ge \limsup_{k \to \infty} |\nu(E_k)| \ge \epsilon > 0 \]

This leads to a severe contradiction! Thus, the assumption is false and the original proposition is proven.

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2. Lebesgue-Radon-Nikodym Theorem

In many application scenarios, given an integral measure \(d\nu = f d\mu\), we can easily determine that \(\nu \ll \mu\). Conversely, if it is known that \(\nu \ll \mu\), can we find a function \(f\) such that \(d\nu = f d\mu\)?

The Lebesgue-Radon-Nikodym Theorem answers this question perfectly and provides a more general decomposition structure.

Theorem 4.2.2 (Lebesgue-Radon-Nikodym Theorem)

Let \(\mu\) and \(\nu\) be two \(\sigma\)-finite measures on a measurable space \((X, \mathcal{M})\) (\(\nu\) can be a signed measure).

Then, there exists a unique pair of measures \(\lambda\) and \(\rho\) such that:

\[ \nu = \lambda + \rho \]

and satisfying the following two core conditions:

• Singularity: \(\lambda \perp \mu\) (i.e., there exist complementary sets that separate them completely).

• Absolute Continuity: \(\rho \ll \mu\). Furthermore, there exists a \(\mu\)-almost everywhere unique function \(f \in L^1(\mu)\) (if \(\nu\) does not take negative infinity, this is an extended locally integrable function) such that for all measurable sets \(E \in \mathcal{M}\):

\[ \rho(E) = \int_E f \, d\mu \]

• Note: This decomposition is called the Lebesgue decomposition of \(\nu\) with respect to \(\mu\). \(\lambda\) is called the singular part, and \(\rho\) is called the absolutely continuous part. The function \(f\) serves exactly as the "density" between the two.

Core Constructive Proof of the Lebesgue-Radon-Nikodym Theorem (for the finite non-negative case)

For simplicity, assume \(\mu\) and \(\nu\) are finite non-negative measures. We seek \(f\) by constructing the largest "under-function."

Define the following family of functions:

\[ \mathcal{F} = \left\{ g : X \to [0, +\infty] \text{ is a measurable function } \mid \int_E g \, d\mu \le \nu(E), \forall E \in \mathcal{M} \right\} \]

Clearly, the constant function \(g = 0\) is in \(\mathcal{F}\), so \(\mathcal{F}\) is non-empty. Moreover, if \(g_1, g_2 \in \mathcal{F}\), it can be proven that \(\max(g_1, g_2) \in \mathcal{F}\).

Let \(M = \sup_{g \in \mathcal{F}} \int_X g \, d\mu\). Since \(\nu(X) < \infty\), we have \(M < \infty\). We can obtain a sequence \(\{g_n\} \subset \mathcal{F}\) such that \(\int g_n d\mu \to M\). Let \(f_n = \max(g_1, \dots, g_n)\), then \(\{f_n\}\) is a monotonically increasing sequence of functions, and \(f_n \in \mathcal{F}\).

According to the Monotone Convergence Theorem (MCT), let \(f = \lim_{n \to \infty} f_n = \sup_n f_n\), then \(f \in \mathcal{F}\) and:

\[ \int_X f \, d\mu = M \]

Now, we define the remaining measure part \(\lambda\) as:

\[ \lambda(E) = \nu(E) - \int_E f \, d\mu \]

From \(f \in \mathcal{F}\), it follows that \(\lambda\) is a non-negative measure. We now only need to prove \(\lambda \perp \mu\). If \(\lambda\) were not singular to \(\mu\), then according to measure theory techniques, we could necessarily find a tiny perturbation \(\epsilon \mu \le \lambda\) on some set of positive measure, such that \(f + \epsilon \chi_A\) would still belong to \(\mathcal{F}\). However, this would result in \(\int (f + \epsilon \chi_A) d\mu = M + \epsilon \mu(A) > M\), contradicting the fact that \(M\) is the supremum!

Therefore, \(\lambda\) must be singular with respect to \(\mu\), proving the non-negative case of the theorem. Signed measures or \(\sigma\)-finite measures can be reduced to this case through the Hahn decomposition and space partitioning.

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3. Radon-Nikodym Theorem and Derivatives

As a direct corollary of the Lebesgue decomposition theorem, when it is known in advance that \(\nu \ll \mu\), the singular part \(\lambda\) collapses to 0.

Theorem 4.2.3 (Radon-Nikodym Theorem)

Let \(\mu, \nu\) be \(\sigma\)-finite measures, and \(\nu \ll \mu\).

Then there must exist a \(\mu\)-almost everywhere unique function \(f\) such that for any measurable set \(E \in \mathcal{M}\):

\[ \nu(E) = \int_E f \, d\mu \]

This miraculous function \(f\) is called the Radon-Nikodym derivative of \(\nu\) with respect to \(\mu\), usually denoted by the fractional symbol:

\[ f = \frac{d\nu}{d\mu} \]

The Radon-Nikodym derivative perfectly simulates the derivative chain rule from calculus within integral operations.

Proposition 4.2.1 (Properties of the Radon-Nikodym Derivative)

Assume all measures are \(\sigma\)-finite measures.

• Integration Rule: If \(g \in L^1(\nu)\), then \(g \frac{d\nu}{d\mu} \in L^1(\mu)\), and there is an intuitive change-of-variable formula:

\[ \int_X g \, d\nu = \int_X g \frac{d\nu}{d\mu} \, d\mu \]

• Chain Rule: If \(\nu \ll \mu\) and \(\mu \ll \lambda\), then \(\nu \ll \lambda\), and almost everywhere we have:

\[ \frac{d\nu}{d\lambda} = \frac{d\nu}{d\mu} \frac{d\mu}{d\lambda} \]

• Reciprocal Rule: If \(\mu \ll \nu\) and \(\nu \ll \mu\) (i.e., they are equivalent, possessing identical null sets), then almost everywhere we have:

\[ \frac{d\nu}{d\mu} = \left( \frac{d\mu}{d\nu} \right)^{-1} \]

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4. Complex Measures and Polar Decomposition

The values of a signed measure are restricted to real numbers. If we allow measures to take complex values, we obtain complex measures, which are very important in functional analysis.

Definition 4.2.2 (Complex Measure)

Let \((X, \mathcal{M})\) be a measurable space. A set function \(\nu: \mathcal{M} \to \mathbb{C}\) is called a complex measure if for any sequence of disjoint measurable sets \(\{E_j\}\) it satisfies:

\[ \nu\left(\bigcup_{j=1}^\infty E_j\right) = \sum_{j=1}^\infty \nu(E_j) \]

Note: To ensure the sum of a complex series is independent of the order of terms, the series must be absolutely convergent. This means complex measures naturally take only finite values and cannot take \(\infty\).

Any complex measure \(\nu\) can be uniquely decomposed into real and imaginary parts: \(\nu = \nu_r + i \nu_i\), where \(\nu_r\) and \(\nu_i\) are both finite signed measures.

Definition 4.2.3 (Total Variation)

For a complex measure \(\nu\), define its total variation measure \(|\nu|\) as:

\[ |\nu|(E) = \sup \sum_{j=1}^n |\nu(E_j)| \]

where the supremum is taken over all finite measurable partitions \(\{E_j\}_{j=1}^n\) of the set \(E\). \(|\nu|\) is also a finite non-negative measure.

With the support of the Radon-Nikodym theorem, we can perform an exquisite "polar decomposition" of a complex measure, much like representing a complex number in polar form \(z = r e^{i\theta}\).

Theorem 4.2.4 (Polar Decomposition of Complex Measures)

Let \(\nu\) be a complex measure. Since from the definition of total variation it is clear that \(\nu \ll |\nu|\), by the Radon-Nikodym theorem, there must exist a complex measurable function \(h = \frac{d\nu}{d|\nu|}\) such that:

\[ d\nu = h \, d|\nu| \]

More importantly, this function \(h\) satisfies, \(|\nu|\)-almost everywhere:

\[ |h(x)| = 1 \]

Using polar decomposition, integration with respect to a complex measure can be strictly defined as a conversion to integration with respect to its total variation measure (a non-negative measure). If \(f \in L^1(|\nu|)\), we define:

\[ \int_X f \, d\nu = \int_X f \cdot h \, d|\nu| \]

From this definition, we can very directly derive the inequality bound for the absolute value of a complex integral:

\[ \left| \int_X f \, d\nu \right| \le \int_X |f| \, d|\nu| \]

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