0
$\begingroup$

Say $X$ is a set with endowed with an algebra $\mathcal{R} \subseteq 2^X$ and a $\sigma$-additive measure $\mu : \mathcal{R} \to \mathbb{R}$. We can apply the Lebesgue extension process to define an outer measure $\mu^* : 2^X \to \mathbb{R}$ and then define Lebesgue-measurable sets. It's well known that a set $A \subseteq X$ is measurable if and only if for all $\epsilon > 0$ there exists some $B \in \mathcal{R}$ such that $\mu^*(A \triangle B) < \epsilon$.

Say we have a finite collection of pairwise disjoint measurable sets $A_1,\ldots,A_n \subseteq X$. For any $\epsilon$, I'd like to be able to find the approximations $B_i \in \mathcal{R}$ so that they're pairwise disjoint as well. Is this possible?

$\endgroup$

1 Answer 1

1
$\begingroup$

Yes.

Take any $B_i \in \mathcal{R}$ such that $\mu^*(A_i \triangle B_i) \leqslant \varepsilon$ and put $B_i' = B_i \setminus \bigcup \limits_{j<i} B_j$. Then $B_i' \in \mathcal{R}$ are the desired approximations: they are obviously disjoint and

$$\mu^*(B_i \triangle B_i') = \mu^* \left( B_i \cap \bigcup_{j<i} B_j \right) \leqslant \sum_{j<i} \mu^*(B_i \cap B_j).$$

But

$$\mu^*(B_i \cap B_j) \leqslant \mu(B_i \triangle A_i) + \mu^*(A_i \cap A_j) + \mu^*(A_j \triangle B_j) \leqslant 2 \varepsilon $$

hence

$$\mu^*(B_i \triangle B_i') \leqslant (i-1) \cdot 2 \varepsilon \leqslant 2(n-1) \varepsilon.$$

So finally

$$\mu^*(A_i \triangle B_i') \leqslant \mu^*(A_i \triangle B_i) + \mu^*(B_i \triangle B_i') \leqslant \varepsilon + 2(n-1)\varepsilon = (2n-1)\varepsilon.$$

It suffices to replace $\varepsilon$ with $\frac{\varepsilon}{2n-1}$.

$\endgroup$
1
  • 1
    $\begingroup$ Basically - just make them disjoint. $\endgroup$
    – Adayah
    Sep 26, 2021 at 17:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .