📂Measure Theory

Hahn Decomposition Theorem

Theorem¹

• (a) Let $\nu$ be a signed measure defined on a measurable space $(X, \mathcal{E})$. Then there exist a positive set $P$ and a negative set $N$ for $\nu$, satisfying the following:

  $$P \cup N=X \quad \text{and} \quad P \cap N =\varnothing$$

  Such a $X=P \cup N$ is called a Hahn decomposition for $\nu$.

• (b) Let $P^{\prime}, N^{\prime}$ be another pair of sets satisfying (a). Then the following sets are null sets for $\nu$:

  $$(P-P^{\prime}) \cup (P^{\prime}-P)=(N-N^{\prime}) \cup (N^{\prime}-N)$$

  This can be denoted using the symmetric difference symbol as follows:

  $$P\Delta P^{\prime}=N\Delta N^{\prime}$$

Explanation

(a) For any given measurable space, it is possible to separate the set $X$ into positive and negative sets with respect to $\nu$ defined on the measurable space.

(b) As stated above, even if there are multiple ways to divide the set $X$, essentially, there is no difference. $P$ and $P^{\prime}$, $N$ and $N^{\prime}$ always differ by only a null set, so they may be different from the set perspective but are the same from the measure perspective.

Proof

Strategies: The** proof of this theorem itself is not very difficult, but the flow of the proof is not trivial, so I will explain it concretely before starting. First, define some positive set $P$. Then define $N$ as $N:=X-P$. If $N$ is a negative set, then the proof for (a) is complete. Before proving that $N$ is a negative set, we will verify that $N$ has two properties as defined above. Finally, we will use proof by contradiction. Assuming $N$ is not a negative set, we will complete the proof by showing that a contradiction arises using the two properties.

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Without loss of generality, assume that $\nu$ does not take the value $+\infty$. For the other case, the same proof applies by considering $-\nu$. Let $C$ be the collection of all positive sets in $\mathcal{E}$. Then, by assumption, $\nu$ does not take the value $+\infty$, so there exists $M$ defined as below:

$$M:=\sup \limits_{P \in C } \nu (P) < \infty$$

Now, we can show that there exists a maximizer $P$ satisfying $\nu (P)=M$. Consider the following maximizing sequence $\left\{ P_{j} \right\}$:

$$\lim \limits_{j \rightarrow \infty} \nu (P_{j})=M$$

Since there is no containment relationship between $P_{j}$’s, consider the following $\tilde{P_{j}}$:

$$\tilde{P_{j}} :=\bigcup \limits_{k=1}^j P_{k}$$

Then, $\nu (P_{j}) \le \nu (\tilde{P_{j}}) \le M$, so $\left\{ \tilde{P_{j}} \right\}$ is a maximizing sequence. Also, it is obvious that $\tilde{P_{1}} \subset \tilde{P_2}\subset \cdots$ by definition. Now, define $P$ as follows:

$$P := \bigcup \limits_{j=1}^\infty \tilde{P_{j}}$$

Then, the following holds:

$$\nu (P)=\lim \limits_{j\rightarrow \infty} \nu (\tilde{P_{j}})=M$$

Thus, we have shown the existence of a maximizer satisfying $\nu (P)=M$. Moreover, since $P$ is the countable sum of positive sets, it is a positive set. In fact, such $P$ and $N:=X-P$ are the decomposition mentioned in the theorem. The process of proving that $N$ is such a negative set remains. Let’s now consider $N:=X \setminus P$. As explained above, the proof ends if we show that $N$ is a negative set. First, let’s prove that such $N$ has the following two properties:

• Claim 1 $N$ does not contain any positive set with a measure value greater than $0$. In other words, it does not contain any non-null positive set. That is, if $\nu (E)>0$ and $E$ is a positive set, then $E \not \subset N$.

  Note that it is possible for a set $E \subset N$ to exist that is neither a positive nor a negative set. In other words, a subset of $N$ can be 1. a null set, 2. a negative set, or 3. a set that is neither positive nor negative.

  • Proof

    Suppose $E\subset N$ is a positive set and $\nu (E) >0$. Then, by the definition of $N$, $E$ and $P$ are disjoint sets. Therefore, the following holds:

    $$\nu (P \cup E)=\nu (P)+\nu (E)$$

    However, since $\nu (P)=M$, the following holds:

    $$\nu (P \cup E)=\nu (P)+\nu (E)>M$$

    But this contradicts the assumption that $M=\sup \nu (F)\ \forall F\in \mathcal{E}$. Therefore, there does not exist $E \subset N$ such that $E$ is a positive set and $\nu (E)>0$.

• Claim 2 If $A \subset N$ and $\nu (A)>0$, then there exists $B \subset A$ such that $\nu (B) > \nu (A)$.

  • Proof

    Let’s say $A \subset N$ and $\nu (A)>0$. Then, by Claim 1, $A$ is not a positive set. Therefore, it is neither a null set nor a positive set. Hence, there exists $C$ satisfying the following conditions²:

    $$C \subset A,\ \nu (C) <0$$

    Now, let’s define $B:=A-C$. Then, the following holds:

    $$\nu (A)=\nu (B)+\nu (C) < \nu (B)$$

Now, let’s assume $N$ is not a negative set. By showing that a contradiction arises using the above two properties, we prove that $N$ is a negative set.

• Part 1.

  Let $\left\{ A_{j} \right\}$ be a sequence of subsets of $N$. Let $\left\{ n_{j} \right\}$ be a sequence of natural numbers. Assuming $N$ is not a negative set, there exists some $B \subset N$ with $\nu (B) >0$. Let’s say the smallest $n_{j}$ satisfying $\nu (B) > \frac{1}{n_{j}}$ is $n_{1}$, and let’s call such $B$ as $A_{1}$. Since $\nu (B)=\nu (A_{1})>0$, the process done for $N$ can be applied to $A_{1}$ equally.

• Part 2

  Again, there exists some $B\subset A_{1}$ with $\nu (B)>0$, and by Claim 2, $\nu (B) > \nu (A_{1})$. Therefore, there exists a natural number $n$ such that $\nu (B) > \nu (A_{1})+\frac{1}{n}$. Let’s call the smallest such natural number $n_2$, and such $B$ as $A_2$.

• Part 3

  Repeating the same process, $n_{j}$ is the smallest natural number for which there exists some $B \subset A_{j-1}$ satisfying $\nu (B)>\nu (A_{j-1}) + \dfrac{1}{n_{j}}$, and such $B$ is called $A_{j}$. Now, let’s define $A=\bigcap \nolimits_{1}^\infty A_{j}$. Since $\nu$ is assumed not to take the value $+\infty$ and by the property of signed measure $(B)$, the following holds:

  $$\begin{align*} +\infty \gt \nu (A) &= \nu \left(\bigcap \nolimits_{1}^\infty A_{j} \right) \\ &= \lim \limits_{j \rightarrow \infty} \nu (A_{j}) \\ &\ge \lim \limits_{j\rightarrow \infty} \left( \nu (A_{j-1}) +\frac{1}{n_{j}} \right) \\ &\ge \lim \limits_{j\rightarrow \infty} \left( \nu (A_{j-2}) + \frac{1}{n_{j-1}} +\frac{1}{n_{j}} \right) \\ &\vdots \\ &\ge \lim \limits_{j\rightarrow \infty} \left( \nu (A_{1}) + \frac{1}{n_{2}}+\cdots +\frac{1}{n_{j}} \right) \\ &\ge \lim \limits_{j\rightarrow \infty} \left( \frac{1}{n_{1}}+ \frac{1}{n_{2}}+\cdots +\frac{1}{n_{j}} \right) \\ &= \sum \limits_{j=1}^\infty \frac{1}{n_{j}} \end{align*}$$

  Since the series is finite, the limit is $0$.

  $$\lim \limits_{j\rightarrow \infty} \frac{1}{n_{j}} =0$$

  Therefore, we obtain the following:

  $$\begin{equation} \lim \limits_{j\rightarrow \infty} n_{j} =\infty \label{eq1} \end{equation}$$

  However, as seen in Part 1, by Claim 2, there exists some natural number $n$ for which there exists $B \subset A$ satisfying $\nu (B) > \nu (A) +\dfrac{1}{n}$. Then, by the definition of $A$, $A \subset A_{j-1}$, and by Claim 2, the sequence $\left\{ \nu (A_{j}) \right\}$ is increasing. Therefore, since $\nu (A) =\lim \limits_{j \rightarrow \infty} \nu (A_{j})$, $\nu (A) > \nu (A_{j-1})$.

  Additionally, by $(1)$, for sufficiently large $j$, $n_{j} >n$. Therefore, the following holds:

  $$\nu (B) > \nu (A) +\frac{1}{n}>\nu (A_{j-1}) +\frac{1}{n} > \nu (A_{j-1}) +\frac{1}{n_{j}}$$

  But this contradicts the definition of $n_{j}$ and $A_{j}$. Therefore, the assumption that $N$ is not a negative set is wrong. Hence, $N$ is a negative set.

  Let’s consider $P^{\prime}$, $N^{\prime}$ as another decomposition satisfying the theorem. Then, the following holds:

  $$P^{\prime} \cup N^{\prime} =X \quad \text{and} \quad P^{\prime}\cap N^{\prime} =\varnothing$$

  Therefore, we can see that $P-P^{\prime} \subset P$, $P-P^{\prime}\subset N^{\prime}$. Then, $P-P^{\prime}$ is both a positive set and a negative set, which is only possible if it’s a null set, so $P-P^{\prime}$ is $\nu-\mathrm{null}$. Similarly, $\nu -\mathrm{null}$ can be shown for $P^{\prime}-P$, $N-N^{\prime}$, and $N^{\prime}-N$.

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