Measure of the union of the sets of a numerable family of measurable sets

Question

I'm working with some measure exercises . I've tried to get a formula to know the measure of the union of n sets ( finite number ), I think I have it, proved by induction, I wonder if there is also a formula to measure the infinite union of all the sets in a numerable family or if that can't be simply measured or if there doesn't exist a formula. Thanks.

Answer 1

Welcome to MSE!

I'm going to link a related question that doesn't look like it ever got a full answer, in case it's of interest to you.

The result you're looking for is called inclusion-exclusion, and because the relevant sum is alternating it gives rise to a family of inequalities called Bonferroni Inequalities. This is another place where knowing the name of something lets you find endless information about it. I suggest you google "countable inclusion exclusion principle" or something similar now that you know!

For convenience, though, here is one result in this vein, which you can find in the paper "Exact Conditions for Countable Inclusion-Exclusion Identity and Extensions" by Friedland and Krop (see here).

$$\mu \left ( \bigcup_{i \in \mathbb{N}} A_i \right ) = \sum_{k \in \mathbb{N}} (-1)^{k-1} \sum_{1 \leq i_1 \leq \ldots \leq i_k} \mu \left ( A_{i_1} \cap \ldots \cap A_{i_k} \right )$$

Then the authors give necessary conditions for this to converge (after all, it's an infinite sum). First let

$$ S_k \triangleq \sum_{1 \leq i_1 \leq \ldots \leq i_k} \mu \left ( A_{i_1} \cap \ldots \cap A_{i_k} \right )$$

Then the paper says

[The result] holds if $S_k$, $k \in \mathbb{N}$ is a sequence of nonnegative numbers that converge exponentially to zero:

$$ S_k \in [0, \infty) \text{ for } k \in \mathbb{N} \text{ and } \limsup_{k \to \infty} S_k^{\frac{1}{k}} < 1$$

You can check out the original paper for more information.

---

I hope this helps ^_^

Answer 2

If $A=\{A_n:n\in \Bbb N\}$ is a family of measurable sets then $\mu(\cup A)=\sum_{n\in \Bbb N}\mu (B_n)$ where $B_1=A_1$ and $B_{n+1}=A_{n+1}\setminus (\cup_{j=1}^n A_j).$