如何证明σ-代数上σ-可加测度μ≪ν等价于lim(ν(S)→0)μ(S)=0？

Definitions #

First, let's formalize the two definitions we're working with to eliminate any ambiguity:

• Definition 1 (ε-δ Absolute Continuity): Given a σ-algebra $\mathcal{S} \subseteq 2^\Omega$, a normed $\mathbb{R}$-vector space $E$, and σ-additive maps $\mu: \mathcal{S} \to E$, $\nu: \mathcal{S} \to [0, +\infty)$, we write $\mu \ll \nu$ if:

  $\forall \varepsilon > 0: \exists \delta > 0: \forall S \in \mathcal{S}: \nu(S) < \delta \implies |\mu(S)|_E < \varepsilon$ (Condition (1))

• Definition 2 (Null-Set Absolute Continuity): $\mu$ is absolutely continuous with respect to $\nu$ if:

  $\forall S \in \mathcal{S}: \nu(S) = 0 \implies \mu(S) = 0$ (Condition (2))

Proof of Equivalence #

We'll verify both directions of the equivalence to tie the two definitions together.

Direction 1: (1) ⇒ (2) #

This implication is straightforward. Suppose Condition (1) holds, and let $S \in \mathcal{S}$ be any set where $\nu(S) = 0$. For any $\varepsilon > 0$, since $\nu(S) = 0$ is less than the $\delta > 0$ guaranteed by Condition (1), we must have $|\mu(S)|_E < \varepsilon$.

Since $\varepsilon$ can be made arbitrarily small, the only element of $E$ with a norm smaller than every positive $\varepsilon$ is the zero vector. Thus $\mu(S) = 0$, so Condition (2) holds.

Direction 2: (2) ⇒ (1) #

We'll use proof by contradiction here. Suppose Condition (2) holds, but Condition (1) does not hold. That means there exists some fixed $\varepsilon_0 > 0$ such that for every $\delta > 0$, there's a set $S \in \mathcal{S}$ where $\nu(S) < \delta$ but $|\mu(S)|_E \geq \varepsilon_0$.

Let's take $\delta_n = 1/2^n$ for each $n \in \mathbb{N}$. By our assumption, we can find a sequence of sets ${S_n}_{n=1}^\infty \subseteq \mathcal{S}$ where:

• $\nu(S_n) < 1/2^n$ for all $n$
• $|\mu(S_n)|_E \geq \varepsilon_0$ for all $n$

Now define the lim sup of these sets:
$$T = \limsup_{n \to \infty} S_n = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty S_k$$

First, calculate $\nu(T)$. For each $n$, $\nu\left(\bigcup_{k=n}^\infty S_k\right) \leq \sum_{k=n}^\infty \nu(S_k) < \sum_{k=n}^\infty 1/2^k = 1/2^{n-1}$. Since ${ \bigcup_{k=n}^\infty S_k }$ is a decreasing sequence of sets (as $n$ increases, we take smaller unions) and $\nu\left(\bigcup_{k=1}^\infty S_k\right) \leq 1 < \infty$, σ-additivity of $\nu$ gives us:
$$\nu(T) = \lim_{n \to \infty} \nu\left(\bigcup_{k=n}^\infty S_k\right) = 0$$

By Condition (2), this means $\mu(T) = 0$. Now, since ${ \bigcup_{k=n}^\infty S_k }$ is a decreasing sequence converging to $T$, σ-additivity of the $E$-valued measure $\mu$ tells us:
$$\lim_{n \to \infty} \mu\left(\bigcup_{k=n}^\infty S_k\right) = \mu(T) = 0$$
Taking norms, this means $\lim_{n \to \infty} \left| \mu\left(\bigcup_{k=n}^\infty S_k\right) \right|_E = 0$.

Now apply the reverse triangle inequality. For each $n$, $S_n \subseteq \bigcup_{k=n}^\infty S_k$, so:
$$\left| \mu\left(\bigcup_{k=n}^\infty S_k\right) \right|E = \left| \mu(S_n) + \mu\left( \bigcup{k=n}^\infty S_k \setminus S_n \right) \right|_E$$
Rearranging the reverse triangle inequality gives:
$$\left| \mu(S_n) \right|E - \left| \mu\left( \bigcup{k=n+1}^\infty S_k \right) \right|E \leq \left| \mu\left(\bigcup{k=n}^\infty S_k\right) \right|_E$$

Let $a_n = \left| \mu\left(\bigcup_{k=n}^\infty S_k\right) \right|E$. Substituting our known $|\mu(S_n)|E \geq \varepsilon_0$, we get:
$$\varepsilon_0 - a{n+1} \leq a_n$$
Or rearranged:
$$a_n + a{n+1} \geq \varepsilon_0$$

But we already know $a_n \to 0$ as $n \to \infty$. Taking the limit of both sides gives $0 + 0 \geq \varepsilon_0$, which is a contradiction (since $\varepsilon_0 > 0$).

This contradiction means our initial assumption (that (1) fails while (2) holds) is wrong. So Condition (2) implies Condition (1).

内容的提问来源于stack exchange，提问作者0xbadf00d