1
$\begingroup$

I am looking at the construction of the Lebesgue measure, and the following doubt came up. When defining the outer Lebesgue measure, we let $$m\left(\bigcup_{k \in \mathcal{K}}I_k\right) = \sum_{k \in \mathcal{K}} m(I_k)$$ for any countable pairwise disjoint collection of intervals $\{I_k \mid k \in \mathcal{K}\}$, thus satisfying the countable additivity requirement.

Now I want to argue that this definition also works for any countable collection of intervals, not just pairwise disjoint ones. My understanding is that the standard construction is the following. Given a countable collection $\mathcal{A} = \{A_1, A_2, A_3, \dots\}$, we can construct $\mathcal{B} = \{B_1, B_2, B_3, \dots\}$ by $$B_n = A_n \cap \left(\bigcap^{n - 1}_{i = 1} (A_i)^c\right).$$ Then we have $\bigcup_i A_i = \bigcup_i B_i$. So back to the outer measure, if we have a countable collection $\mathcal{I}$ of intervals, then we can use the previous procedure to construct a countable pairwise disjoint collection $\mathcal{I}'$ such that $$m\left(\bigcup \mathcal{I}\right) = m\left(\bigcup \mathcal{I}'\right).$$

My question is whether my understanding of this is correct, and under which conditions the construction holds. I guess we have to have closure under intersection for the construction to hold, but is there anything else?

$\endgroup$
2
+50
$\begingroup$

The property you use here is the following: If $A$ and $B$ are intervals, then the difference $A\setminus B$ can be written as the disjoint union of a finite number of intervals (countable would suffice), so $A \setminus B = \biguplus_{i=1}^n I_i$. Iterating this property, we see that each $B_n = (\cdots(A_n \setminus A_1) \setminus \cdots) \setminus A_{n-1}$ is a finite disjoint union of intervals, and hence $\bigcup_n B_n$ equals a disjoint union of a countable number of intervals.

This property, together with closure under intersection has a name:

Definition A set of sets $\mathcal S \subseteq \mathfrak P(X)$ is called semiring on $X$, if all of the following holds:

  1. $\emptyset \in \mathcal S$
  2. $\mathcal S$ is closed under finite intersection.
  3. If $A, B \in \mathcal S$, then $A \setminus B$ is the finite union of disjoint elements from $\mathcal S$.

So, for your construction to work, you need a semiring of sets.

$\endgroup$
  • $\begingroup$ So just to make absolutely sure: Whenever I'm working in a semiring, every countable collection of sets $A_i$ can be turned into an equivalent collection of sets $B_i$ (in the sense that $\bigcup_i A_i = \bigcup_i B_i$) where the $B_i$s are pairwise disjoint. And since every $\sigma$-algebra is also a semiring, this construction works in particular for $\sigma$-algebras. Is this correct? $\endgroup$ – mrp Oct 10 '15 at 7:35
  • $\begingroup$ Yes, that's correct $\endgroup$ – martini Oct 10 '15 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.