Radon-Nikodým Derivative and Lebesgue Decomposition Theorem

Throughout this note, we work on a measurable space \((\Omega, \mathcal{F})\). All involved functions are assumed to be measurable.

Definition [signed measure]. A set function \(\varphi: \mathcal{F} \to \mathbb{R} \cup \{\infty\}\) is called a signed measure if: 1) \(\varphi(\emptyset) = 0\); 2) if \(E = \bigcup_{n=1}^\infty E_n\) is a disjointed union and \(E_n \in \mathcal{F}\) for all \(n\), then^[1] \(\varphi(E) = \sum_{n=1}^\infty \varphi(E_n)\).

Remark. Since we do not allow \(\varphi\) takes negative infinity, it is thus impossible that a finite-measured set contains an infinite-measured set. It is also impossible a finite-measured set contains a sequence of sets whose measures approach to infinity. That is, if \(E \in \mathcal{F}\) has \(\varphi(E) < \infty\), then^[2]

\begin{equation*} \sup\{ \varphi(F) \mid F \in \mathcal{F}, ~ F \subset E\} = M < \infty. \end{equation*}

Definition [measure]. A measure is a signed measure that takes value in \([0, \infty]\). A measure \(\mu\) is called finite if \(\mu(\Omega) < \infty\). A measure \(\mu\) is called σ-finite if there exists a sequence \(\{E_n\}\) such that: 1) \(\mu(E_i) < \infty\) for all \(i\); 2) \(\bigcup E_n = \Omega\).

Remark. We will see that a signed measure can always be written as the difference of two measures.

Definition [positive and negative sets]. Let \(\varphi\) be a signed measure. A measurable set \(A\) is called positive (under \(\varphi\)) if every measurable \(E \subset A\) has \(\varphi(E) \geq 0\). A measurable set \(B\) is called negative (under \(\varphi\)) if every measurable \(E \subset B\) has \(\varphi(E) \leq 0\).

Remark. The concepts of postive and negative sets will be used in the Hahn decomposition theorem, which naturally leads to the Jordan decomposition theorem. These two theorems are helpful for understanding our main topic—the Lebesgue decomposition theorem and Radon-Nikodým theorem.

Definition [absolute continuity]. A measure \(\nu\) is called absolutely continous w.r.t. another measure \(\mu\) if

\begin{equation} \label{eq:absolute-continuity} \forall A ,~ \mu(A) = 0 ~\Rightarrow ~ \nu(A) = 0. \end{equation}

This relation is denoted by \(\nu \ll \mu\). In this case, \(\nu\) is also said to be dominated by \(\mu\). These two measures are called equivalent if \(\nu \ll \mu\) and \(\mu \ll \nu\).

Remark. Let \(\mu\) be a measure and \(f\) be function with \(\int f^-\,d\mu < \infty\). Then, the set function

\begin{equation} \label{eq:integral-as-set-function} \nu(A) := \int_A f\,d\mu \end{equation}

is a signed measure^[3]. If \(f\) is nonnegative, then \(\nu\) becomes a measure. In this case, we can further show that^[4] \(\nu \ll \mu\). The nonnegative function \(f\) is called the density of \(\nu\) w.r.t. \(\mu\). The Radon-Nikodým theorem asserts that such a density function can always be found if \(\nu \ll \mu\), assuming both measures are σ-finite.

Definition [singularity]. A measure \(\nu\) is called singular w.r.t. another measure \(\mu\) if

\begin{equation} \label{eq:singularity} \exists N,~ \nu(N) = \mu(N^c) = 0. \end{equation}

This relation is denoted by \(\nu \perp \mu\). In this case, these two measures are also said to be mutually singular, since this relation is symmetric.

Remark. As an example, the Dirac measure centered at 0 is singular w.r.t. the Lebesgue measure.

In most cases, a measure \(\nu\) may neither be absolutely continuous nor singular to another measure \(\mu\). Nevertheless, the Lebesgue decomposition theorem tells us that \(\nu\) can always be split into two parts—one is absolutely continuous to \(\mu\) and the other one is singular to \(\mu\).

We first state two useful decomposition theorems.

Theorem [Hahn decomposition]. If \(\varphi\) is a signed measure, then there exist a postive set \(A\) and a negative set \(B\) under \(\varphi\), so that \(A \cup B = \Omega\) and \(A \cap B = \emptyset\).

Remark. The Hahn decomposition is unique up to \(\varphi\)-null sets. A measurable set \(N\) is called a \(\varphi\)-null set if any measurable subset \(E \subset N\) has \(\varphi(E) = 0\). Clearly, a measurable set is \(\varphi\)-null set if and only it is both positive and negative under \(\varphi\). Suppose \(\Omega = A_1 \cup B_1 = A_2 \cup B_2\) are two Hahn decompositions. We can show that^[5] both \(A_1 \cap A_2^c\) and \(A_2 \cap A_1^c\) are \(\varphi\)-null sets.

Theorem [Jordan decomposition]. If \(\varphi\) is a signed measure, then there exist a pair of mutually singular measures \(\varphi_+\) and \(\varphi_-\), so that \(\varphi = \varphi_+ - \varphi_-\). Moreover, such a pair is unique.

Remark. We can further prove that^[6] \(\varphi_+(E) = \sup\{\varphi(F) \mid F \subset E,~F \in \mathcal{F}\}\). This can be viewed as an explicit construction of the Jordan decomposition.

Proposition [construction of absolutely continuous measures and singular measures]. Let \(\nu_1,\nu_2,\mu\) be measures on \((\Omega, \mathcal{F})\). Then^[7]

1. \(\nu_1 \ll \mu\) and \(\nu_2 \ll \mu\) imply \((\nu_1 + \nu_2) \ll \mu\).
2. \(\nu_1 \ll \mu\) and \(\nu_2 \perp \mu\) imply \(\nu_1 \perp \nu_2\).
3. \(\nu_1 \perp \mu\) and \(\nu_2 \perp \mu\) imply \((\nu_1 + \nu_2) \perp \mu\).

Remark. It is not necessarily true^[8] that \(\nu_1 \perp \mu\) and \(\mu \ll \nu_2\) imply \(\nu_1 \perp \nu_2\). It is also not necessarily true^[9] that \(\nu_1 \perp \mu_1\) and \(\nu_2 \perp \mu_2\) imply \((\nu_1 + \nu_2) \perp (\mu_1 + \mu_2)\). However, it is always true that \(\nu_1 \ll \mu_1\) and \(\nu_2 \ll \mu_2\) imply \((\nu_1 + \nu_2) \ll (\mu_1 + \mu_2)\).

The above implications will become more intuitive once we are familiar with the Lebesgue decomposition theorem and Radon-Nikodým theorem.

Theorem [Lebesgue decomposition]. Let \(\mu\) and \(\nu\) be two σ-finite measures. Then, there exists a decomposition \(\nu = \nu^\circ + \nu^\perp\), so that \(\nu^\circ \ll \mu\) and \(\nu^\perp \perp \mu\). Moreover, such a decomposition is unique.

Remark. The two parts \(\nu^\circ\) and \(v^\perp\) are mutually singular, i.e., \(v^\circ \perp v^\perp\).

Theorem [Radon-Nikodým]. Let \(\mu\) and \(\nu\) be two σ-finite measures and \(\nu \ll \mu\). Then, there exists a nonnegative \(f\) such that \(\nu(A) = \int_A f\,d\mu\) for all \(A \in \mathcal{F}\). The function \(f\) is called Radon-Nikodým derivative, denoted by \(d\nu / d\mu\). This derivative is unique in the sense that two such versions are μ-a.e. equal.

Remark. The reverse direction trivially holds. That is, if \(\mu\) is a finite measure (σ-finite measure respectively) and \(f\) is nonnegative, then the set function \(\nu(A) := \int_A f\,d\mu\) is a finite measure (σ-finite measure respectively).

To prove the Hahn decomposition theorem, we need a lemma.

Lemma [a negative measured set must contain a negative set]. Let \(\varphi\) be a signed measure. If \(E \in \mathcal{F}\) and \(\varphi(E) < 0\), then there must be a negative set \(F \subset E\) so that \(\varphi(F) < 0\).

Proof. If \(E\) is negative, then choose \(F = E\). If \(E\) is not negative, then, we can find \(A_1 \subset E\) such that \(\varphi(A_1) > 0\). Moreover, we can^[10] take \(A_1\) so that \(\varphi(A_1) > \varphi(D)\) holds for any \(D \subset E - A_1\).

Let \(E_1 = E - A_1\). If \(E_1\) is negative, then choose \(F = E_1\). If \(E_1\) is not negative, then we can find \(A_2 \subset E_1\) such that \(\varphi(A_2) > 0\). Similarly, we can take \(A_2\) so that \(\varphi(A_2) > \varphi(D)\) for any \(D \subset E_1 - A_2\).

Let \(E_2 = E_1 - A_2\). If \(E_2\) is negative, then choose \(F = E_2\). If \(E_2\) is not negative, then we can take \(A_3 \subset E_2\) such that \(\varphi(A_3) > 0\). Similarly, we can take \(A_3\) so that \(\varphi(A_3) > \varphi(D)\) for any \(D \subset E_2 - A_3\).

Continue the above process. If \(E_n\) is not negative for all \(n \geq 1\), then there are a decreasing sequence \(\{E_n\}_{n=1}^\infty\) and a disjointed union \(A := \bigcup_n A_n \subset E\). Since \(\varphi(E) < \infty\), we have \(\varphi(A) < \infty\) (see the remark after Definition [signed measure]). Thus, the series \(\sum_n \varphi(A_n)\) converges absolutely and the limit is \(\varphi(A)\). We conclude that \(\varphi(A_n) \to 0\).

Let \(E_\infty := E - A = \bigcap_n E_n\). By our construction of \(A\), we must have \(E_\infty\) is negative. To see this, note that for any \(n\) and \(D \subset E_n - A_{n+1}=E_{n+1}\), there must be \(\varphi(D) < \varphi(A_{n+1})\). Let \(B \subset E_\infty\). we have \(B \subset E_{n+1}\) and \(\varphi(B) < \varphi(A_{n+1})\) for any \(n\). As \(\varphi(A_n) \to 0\), there must be \(\varphi(B) \leq 0\). Hence, we found a negative set \(E_\infty \subset E\) with \(\varphi(E_\infty) = \varphi(E) - \varphi(A) < 0\).

In conclusion, we can always find a negative set \(F \subset E\) so that \(\varphi(F) < 0\).

Q.E.D.

Proof to the Hahn decomposition theorem. Let \(\mathcal{B}\) be the collection of negative sets:

\begin{equation*} \mathcal{B}:=\{ B \in \mathcal{F} \mid B \text{ is negative under } \varphi \}. \end{equation*}

Since \(\emptyset\) is also a negative set, the collection \(\mathcal{B}\) contains at least one element.

Let \(\beta = \inf \{\varphi(B) \mid B \in \mathcal{B}\} \leq 0\). We are going to find a negative set \(D \in \mathcal{B}\) so that \(\beta = \varphi(D)\). Let \(\{B_n\}\) be a sequence in \(\mathcal{B}\) such that \(\varphi(B_n) \downarrow \beta\). Let \(D = \bigcup_n B_n\). Noting^[11] \(D \in \mathcal{B}\), we have \(\beta \leq \varphi(D) \leq 0\). On the other hand, \(\varphi(D) \leq \varphi(B_n)\) for all \(n\). Taking the limit yields \(\varphi(D) = \beta\). By our definition of signed measures, \(\varphi(D) > -\infty\).

The set \(D^c\) is positive. Otherwise, assume there exists a measurable subset \(E \subset D^c\) such that \(\varphi(E) < 0\). By the above lemma, we can find a negative set \(F \subset E \subset D^c\) so that \(\varphi(F) < 0\). The union \(F \cup D\) is a negative set and \(\varphi(F \cup D) = \varphi(F) + \varphi(D) < \varphi(D)\), which contradicts \(\varphi(D) = \inf \{\varphi(B) \mid B \in \mathcal{B}\}\). Therefore, \(D^c\) is positive and we have found a Hahn decomposition \(\Omega = D^c \cup D\).

Q.E.D.

Proof to the Jordan decomposition theorem. Let \(\Omega = A \cup B\) be a Hahn decomposition, so that \(A\) is positive and \(B\) is negative under \(\varphi\). Define

\begin{equation*} \varphi_+(E):= \varphi(E \cap A), \quad \varphi_-(E) := -\varphi(E \cap B). \end{equation*}

By this definition, \(\varphi_+\) and \(\varphi_-\) are both measures. Moreover, \(\varphi_-\) is a finite measure^[12]. Furthermore, \(\varphi_+\) and \(\varphi_-\) mutually singular and \(\varphi(E) = \varphi_+(E) - \varphi_-(E)\) for all \(E \in \mathcal{F}\).

To prove the uniqueness, let \(\varphi = \phi_+ - \phi_-\) be another decomposition. Since \(\phi_+\) and \(\phi_-\) are mutually singular, let \(D\) be the set such that \(\phi_+(D) = \phi_-(D^c) = 0\). For any \(E \in \mathcal{F}\), we have

\begin{equation*} \begin{aligned} \phi_+(E) &= \phi_+(E \cap D^c) = \varphi(E \cap D^c) + \phi_- (E \cap D^c) = \varphi(E \cap D^c) \\ \phi_-(E) &= \phi_-(E \cap D) = \phi_+(E \cap D) - \varphi(E \cap D) = -\varphi(E \cap D) \\ \varphi(E) &= \phi_+(E) - \phi_-(E) = \phi_+(E \cap D^c) - \phi_-(E \cap D). \end{aligned} \end{equation*}

This shows that \(D\) is a negative set under \(\varphi\) and that \(\Omega = D^c \cup D\) is another Hahn decomposition. For any \(E \in \mathcal{F}\),

\begin{equation*} \phi_+(E) = \varphi(E \cap D^c) = \varphi(E \cap D^c \cap A) + \varphi(E \cap D^c \cap A^c)= \varphi(E \cap D^c \cap A). \end{equation*}

The last equality holds because \(A^c \cap D^c\) is both positive and negative. Similarly, we have

\begin{equation*} \varphi_+(E) = \varphi(E \cap A)= \varphi(E \cap A \cap D^c) + \varphi(E \cap A \cap D)= \varphi(E \cap D^c \cap A). \end{equation*}

The last equality holds because \(A \cap D\) is both positive and negative.

Q.E.D.

To prove the Lebesgue decomposition theorem and Radon-Nikodým theorem, we prove the following core result.

Theorem [Lebesgue-Radon-Nikodým]. Let \(\mu\) and \(\nu\) be two σ-finite measures. Then, there exist a nonnegative function \(f\) and a measure \(\nu^\perp\) that is singular to \(\mu\) so that

\begin{equation*} \nu(A) = \int_A f\,d\mu + \nu^\perp(A), \quad \forall A \in \mathcal{F}. \end{equation*}

Proof. The proof is split into two cases.

Case 1 (when \(\mu\) and \(\nu\) are both finite measures). To construct the density \(f\), we consider the collection

\begin{equation*} \mathscr{G} := \biggl\{g \geq 0 \biggm| \int_A g\,d\mu \leq \nu(A),\quad \forall A \in \mathcal{F}\biggr\}. \end{equation*}

Let \(\vee\) denote the pointwise maximum of functions: \((g_1 \vee g_2)(\omega):=\max(g_1(\omega),g_2(\omega))\). We can show that^[13] \(\mathscr{G}\) is closed under \(\vee\): if \(g_1, g_2 \in \mathscr{G}\), then \(g_1 \vee g_2 \in \mathscr{G}\). Consider the supremum

\begin{equation*} I := \sup \biggl\{ \int g\,d\mu \biggm| g \in \mathscr{G} \biggr\}. \end{equation*}

Take a sequence \(g_n \in \mathscr{G}\) so that \(\int g_n\,d\mu \uparrow I\). Let \(f_n:= g_1 \vee g_2 \vee \cdots \vee g_n\). We have \(f_n \in \mathscr{G}\) and \(\int f_n\,d\mu \geq \int g_n\,d\mu\) for all \(n\). Hence,

\begin{equation*} I \geq \int f_n \,d\mu \geq \int g_n \,d\mu,\quad \forall n. \end{equation*}

Taking \(n \to \infty\) concludes that \(\int f_n\,d\mu \to I\). On the other hand, let \(f\) be the limit function that \(0 \leq f_n \uparrow f\). By the monotone convergence theorem, its integral \(\int f\,d\mu = \lim \int f_n\,d\mu = I\). Again, by the monotone convergence theorem, we have \(f \in \mathscr{G}\).

Since \(\nu\) is finite, the supremum \(I \leq \nu(\Omega) < \infty\). For any \(A \in \mathcal{F}\), the integral \(\int_A f\,d\mu \leq \int f\,d\mu = I < \infty\). Define a measure (it is indeed a measure, not only a signed measure)

\begin{equation*} \nu^\perp(A) := \nu(A) - \int_A f\,d\mu. \end{equation*}

We need only to show that \(\nu^\perp\) is singular to \(\mu\). Pick an arbitrary \(\epsilon > 0\) and let \(\Omega = A_\epsilon \cup B_\epsilon\) be an Hahn decomposition of \(\nu^\perp - \epsilon \mu\) (this is a well-defined signed measure as \(\mu\) is finite). Since \(A_\epsilon\) is positive under \(\nu^\perp - \epsilon \mu\), we have \(\nu^\perp(E \cap A_\epsilon) \geq \epsilon \mu(E \cap A_\epsilon)\) for all \(E \in \mathcal{F}\). Moreover,

\begin{equation*} \begin{aligned} \int_E (f + \epsilon \mathbb{1}_{A_\epsilon})\,d\mu &= \int_E f\,d\mu + \epsilon \mu(E \cap A_\epsilon) \\ &\leq \int_E f\,d\mu + \nu^\perp(E \cap A_\epsilon) \\ &\leq \int_E f\,d\mu + \nu^\perp(E) \\ &= \nu(E). \end{aligned} \end{equation*}

This suggests that \(f + \epsilon 1_{A_\epsilon} \in \mathscr{G}\). Hence,

\begin{equation*} \int f\,d\mu = I \geq \int (f + \epsilon 1_{A_\epsilon})\,d\mu = \int f\,d\mu + \epsilon \mu(A_\epsilon), \end{equation*}

showing that \(\mu(A_\epsilon) = 0\).

Let \(A := \bigcup_n A_{1/n}\) be a μ-null set. For any \(k\), the set \(A^c = \bigcap_n A^c_{1/n} \subset A^c_{1/k}\) is a negative set under \(\nu^\perp - \mu/k\). Thus,

\begin{equation*} \nu^\perp(A^c) \leq \frac{1}{k} \mu(A^c),\quad \forall k. \end{equation*}

Taking \(k \to \infty\) concludes that \(\nu^\perp(A^c) = 0\). Combining with \(\mu(A) = 0\), we have proved that \(\nu^\perp\) and \(\mu\) are mutually singular. This completes the proof of the first case.

Case 2 (when \(\mu\) and \(\nu\) are both σ-finite measures). By the σ-finiteness, there exists a disjointed union \(\Omega = \bigcup_n \Omega_n\) such that \(\mu(\Omega_n)\) and \(\nu(\Omega_n)\) are finite for all \(n\). Define

\begin{equation*} \mu_n(A) := \mu(A \cap \Omega_n), \quad \nu_n(A):= \nu(A \cap \Omega_n). \end{equation*}

By this definition, \(\mu_n \perp \nu_m\) for all \(n \neq m\).

According to the conclusion of Case 1, there exist a nonnegative function \(f_n\) and \(\nu^\perp_n\) such that

\begin{equation*} \nu_n(A) = \int_A f_n\,d\mu_n + \nu_n^\perp(A), \quad \forall A \in \mathcal{F}, \end{equation*}

and \(\nu^\perp_n \perp \mu_n\). Let \(f := \sum_n f_n \cdot \mathbb{1}_{\Omega_n}\) and \(\nu^\perp := \sum_n \nu_n^\perp\). Pick any \(A \in \mathcal{F}\). Then^[14],

\begin{equation*} \begin{aligned} \nu(A) &= \sum_n \nu(A \cap \Omega_n) \\ &= \sum_n \biggl( \int_A f_n\,d\mu_n + \nu_n^\perp(A) \biggr) \\ &= \sum_n \int_A f_n \cdot \mathbb{1}_{\Omega_n}\,d\mu + \nu^\perp(A) \\ &= \int_{A} f\,d\mu + \nu^\perp(A). \end{aligned} \end{equation*}

We need only to show that \(\nu^\perp\) is singular to \(\mu\). For all \(m \neq n\), we have \(\nu_m \perp \mu_n\) and \(\nu_m^\perp \ll \nu_m\), suggesting that \(\nu_m^\perp \perp \mu_n\). By the conclusion of Case 1, there is \(\nu_n^\perp \perp \mu_n\). Hence, the sum \(\bigl(\sum_n \nu_n^\perp\bigr) \perp \mu_n\). Moreover, we have \(\bigl(\sum_n \nu_n^\perp \bigr) \perp \bigl(\sum_n \mu_n\bigr)\), which is exactly \(\nu^\perp \perp \mu\).

Q.E.D.

Proof to the Lebesgue decomposition theorem. The existence of \(\nu^\circ\) and \(\nu^\perp\) has been proved in Theorem [Lebesgue-Radon-Nikodým], we need only to prove the uniqueness.

Let \(\nu = \nu^\circ + \nu^\perp\) and \(\nu = \lambda^\circ + \lambda^\perp\) be two Lebesgue decompositions of \(\nu\) w.r.t. \(\mu\). By the σ-finiteness, there exists a disjointed union \(\Omega = \bigcup_n \Omega_n\) such that \(\mu(\Omega_n)\) and \(\nu(\Omega_n)\) are finite for all \(n\). Define

\begin{equation*} \begin{aligned} \nu_n(A) &:= \nu(A \cap \Omega_n),&\quad \mu_n(A) &:= \mu(A \cap \Omega_n), \\ \nu^\circ_n(A) &:= \nu^\circ(A \cap \Omega_n),&\quad \nu^\perp_n(A) &:= \nu^\perp(A \cap \Omega_n), \\ \lambda^\circ_n(A) &:= \lambda^\circ(A \cap \Omega_n),&\quad \lambda^\perp_n(A) &:= \lambda^\perp(A \cap \Omega_n). \end{aligned} \end{equation*}

By this definition, \(\nu_n = \nu^\circ_n + \nu^\perp_n\) and \(\nu_n = \lambda^\circ_n + \lambda^\perp_n\) are also Lebesgue decompositions of \(\nu_n\) w.r.t. \(\mu_n\).

Consider the signed measure \(\varphi_n := \nu^\circ_n - \lambda^\perp_n\). Note that \(\nu^\circ_n \ll \mu_n\) and \(\lambda^\perp_n \perp \mu_n\), we have \(\nu^\circ_n \perp \lambda^\perp_n\). Therefore, \(\varphi_n = \nu^\circ_n - \lambda^\perp_n\) is a Jordan decomposition. On the other hand, since all these \(n\)-subscripted measures are finite, we have

\begin{equation*} \varphi_n = \nu^\circ_n - \lambda^\perp_n = (\nu_n - \nu^\perp_n) - (\nu_n - \lambda^\circ_n) = \lambda^\circ_n - \nu^\perp_n. \end{equation*}

A similar argument shows that \(\lambda^\circ_n \perp \nu^\perp_n\), so \(\varphi_n = \lambda^\circ_n - \nu^\perp_n\) is another Jordan decomposition. By the uniqueness of the Jordan decomposition, there are \(\nu_n^\circ = \lambda_n^\circ\) and \(\nu_n^\perp = \lambda_n^\perp\). Since this result holds for all \(n\), we must have \(\nu^\circ = \lambda^\circ\) and \(\nu^\perp = \lambda^\perp\).

Q.E.D.

Proof to the Radon-Nikodým Theorem. By Theorem [Lebesgue-Radon-Nikodým], there exist a nonegative function \(f\) and a measure \(\nu^\perp \perp \mu\) such that for any \(A \in \mathcal{F}\),

\begin{equation*} \nu(A) = \nu^\circ(A) + \nu^\perp(A), \qquad \nu^\circ(A):= \int_A f\,d\mu. \end{equation*}

Since \(\nu^\circ \ll \mu\), the decomposition \(\nu = \nu^\circ + \nu^\perp\) is the Lebesgue decomposition of \(\nu\) w.r.t. \(\mu\). By assumption, we also have \(\nu \ll \mu\), implying that \(\nu = \nu + 0\) is another Lebesgue decomposition. The uniqueness of Lebesgue decomposition implies \(\nu^\perp = 0\).

We need only to show the uniqueness of \(f\). Let \(g\) be another nonnegative function such that \(\nu(A) = \int_A g\,d\mu\) for all \(A \in \mathcal{F}\). We are going to show that \(f = g\) holds μ-a.e..

By the σ-finiteness, there exists \(A_n \uparrow \Omega\) and \(\mu(A_n) < \infty\) for all \(n\). Let \(E_n = [f < g < n]\) and \(h_n = (g - f) \cdot \mathbb{1}_{E_n \cap A_n} \geq 0\). By assumption,

\begin{equation*} \int_{E_n \cap A_n} f\,d\mu = \int_{E_n \cap A_n} g\,d\mu \leq n \cdot \mu(A_n) < \infty. \end{equation*}

Therefore, \(\int h_n\,d\mu =0\) and \(h_n = 0\) holds μ-a.e. for all \(n\).

Consider an arbitrary \(n\). Since \(h_n=0\) holds μ-a.e., there exists a μ-null set \(N_n\) such that \(h_n=0\) holds on \(N_n^c\). Because \(h_n > 0\) on \(E_n \cap A_n\), we have \((E_n \cap A_n ) \cap N_n^c = \emptyset\). As a result, \(E_n \cap A_n \subset N_n\) is a μ-null set. Noting \(\bigcup_n (E_n \cap A_n) = \bigcup_n E_n = [f < g]\) concludes that \(f \geq g\) holds μ-a.e..

Similarly, we can prove \(f \leq g\) holds μ-a.e..

Q.E.D.