If $\mu$ is a finite measure and $\nu$ is a signed measure, can we extend the inequality $|\nu|\le C\mu$ from a generator to the whole σ-algebra?

Question

Let

• $(\Omega,\mathcal A,\mu)$ be a finite measure space
• $\mathcal E$ be a semiring with $\sigma(\mathcal E)=\mathcal A$
• $\nu$ be a signed measure on $(\Omega,\mathcal A)$ with $$|\nu(A)|\le C\mu(A)\;\;\;\text{for all }A\in\mathcal E\tag1$$ for some $C\ge0$

Are we able to show that $(1)$ holds for all $A\in\mathcal A$?

If $|\nu|$ (the composition of the absolute value function and $\nu$; not to be confused with the total variation of $\nu$) would be a measure, then $$\mathcal E\ni A\mapsto C\mu(A)-|\nu(A)|\tag2$$ would be a (nonnegative) pre-measure on $\mathcal E$ and hence admit a unique extension to a (nonnegative) measure on $\mathcal A$. The desired claim would follow immediately.

However, $|\nu|$ should fail to be additive and hence this argumentation is not possible. So, what can we do?

Answer 1

I will assume that $\Omega = \bigcup_{n = 1}^{\infty} A_{n}$, where $\{A_{n}\}_{n \in \mathbb{N}} \subseteq \mathscr{E}$. Note that this assumption is necessary for the "and hence admit a unique extension" result the OP is quoting. See, for example, this post.

In what follows, I will assume that $\{A_{n}\}_{n \in \mathbb{N}}$ is pairwise disjoint. To see this is no restriction, observe that if $A_{n} \cap A_{m} \neq \phi$, then observe that $A_{n} \cap A_{m} \in \mathscr{E}$ and $A_{m} \setminus A_{n}$ is a disjoint union of finitely many sets in $\mathscr{E}$. Thus, cutting up the $\{A_{n}\}_{n \in \mathbb{N}}$ as required, we can obtain a sequence $\{\tilde{A}_{n}\}_{n \in \mathbb{N}} \subseteq \mathscr{E}$ that is pairwise disjoint and covers $\Omega$.

Define a collection $\mathscr{B}$ of subsets of $\Omega$ by $$\mathscr{B} = \{A \in \mathcal{A} \, \mid \, |\nu(A)| \leq C \mu(A) \, \, \text{and} \, \, |\nu(\Omega \setminus A)| \leq C \mu(\Omega \setminus A)\}.$$ $\mathscr{B}$ is a $\lambda$-system. Indeed, $\Omega \in \mathscr{B}$ since $$\left|\nu \left(\bigcup_{n = 1}^{\infty} A_{n} \right) \right| \leq \sum_{n = 1}^{\infty} |\nu(A_{n})| \leq C \sum_{n = 1}^{\infty} \mu(A_{n}) = C \mu(\Omega).$$ Notice that this same reasoning implies that if $\{C_{n}\}_{n \in \mathbb{N}} \subseteq \mathscr{B}$ is pairwise disjoint, then $\bigcup_{n = 1}^{\infty} C_{n} \in \mathscr{B}$. Finally, by construction, $A \in \mathscr{B}$ implies $\Omega \setminus A \in \mathscr{B}$.

Since $\mathscr{E}$ is a semi-ring, it is a $\pi$-system. Suppose $A \in \mathscr{E}$. Since $\mathscr{E}$ is a semi-ring, if $A \in \mathscr{E}$ and $L \in \mathbb{N}$, then there exist pairwise disjoint sets $C_{1},C_{2},\dots,C_{N} \in \mathscr{E}$ such that $\A_{L} \setminus A = C_{1} \cup \dots \cup C_{N}$. Therefore, $$|\nu(A_{L} \setminus A)| \leq |\nu(C_{1})| + \dots + |\nu(C_{N})| \leq C \mu(A_{L} \setminus A).$$ Summing over $L$ (and appealing to the fact that $\{A_{L}\}_{L \in \mathbb{N}}$ is disjoint), we obtain $$|\nu(\Omega \setminus A)| \leq \sum_{L = 1}^{\infty} |\nu(A_{L} \setminus A)| \leq C \sum_{L = 1}^{\infty} \mu(A_{L} \setminus A) = C \mu(\Omega \setminus A).$$ Since $|\nu(A)| \leq C \mu(A)$ by hypothesis, this completes the proof that $A \in \mathscr{B}$. Therefore, $\mathscr{E} \subseteq \mathscr{B}$.

$\mathscr{E}$ is a $\pi$-system contained in the $\lambda$-system $\mathscr{B}$. Therefore, by Dynkin's Theorem, $\mathcal{A} = \sigma(\mathscr{E}) \subseteq \mathscr{B}$.

Answer 2

By the basic approximation result of measure theory (ref. Halmos), for any $A \in \mathcal A$ we can find a sequence $\{A_n\}$ in the semiring such that $\mu (A \Delta A_n)$ tends to 0. Now $|\nu (A_n)-\nu (A_m)| =|\nu (A_n \setminus A_m)- \nu (A_m \setminus A_n)| \leq \mu (A_n \Delta A_m)$ which tends to 0. The Cauchy sequence $\{ \nu (A_n)\}$ is convergent and we can define $\nu (A)$ as its limit. A straight-forward argument shows that $\nu$ is well-defined and additive on $\mathcal A$. Now 1) holds for all A in $\mathcal A$ which also makes teh extension countable additive.