Question regarding a property of the Lebesgue measure on Wikipedia

Question

On wikipedia, the following property of the Lebesgue measure is listed:

If $A$ is a disjoint union of countably many disjoint Lebesgue-measurable sets, then $A$ is itself Lebesgue-measurable and $\mu(A)$ is equal to the sum (or infinite series) of the measures of the involved measurable sets

which I have interpreted as saying that

If $\lbrace \mathcal{X}_i:i\in I\rbrace $ is a family of countably many sets, then let $A$ satisfy $$ A=\bigsqcup_{i\in I}\mathcal{X}_i $$ If $\mathcal{X}_i$ and $\mathcal{X}_j$ are disjoint for any $i,j\in I$, then $A$ is itself Lebesgue-measurable and $\mu(A)$ is equal to the sum (or infinite series) of the measures of the involved measurable sets

Now, I am not certain if this interpretation is correct or not, so any corrections are appreciated, but above all else I find the fact that $A$ has to be the disjoint union rather redundant, since, if every $\mathcal{X}_i$ is disjoint from $\mathcal{X}_j$ for $i\neq j$, then $\mu(\mathcal{X}_i\cup\mathcal{X}_j)=\mu(\mathcal{X}_i\sqcup\mathcal{X}_j)$, so why bother with a disjoint union instead of a standard union? Is there a flaw in my logic?

Answer 1

The reason for the disjoint union is so that we can get a clean formula for the measure without worrying about intersections. A union of sets that have intersections would give us an inclusion-exclusion formula instead, that wouldn't be guaranteed to converge in the infinite case.

Saying "disjoint" twice there is redundant - but we definitely want it at least once., so that we can have the measure of the (disjoint) union be the sum of the measures.

Answer 2

If $A_n, n \in \mathbb{N}$ is a countable family of Lebesgue measurable sets, then $A = \bigcup_{n \in \mathbb{N}} A_n$ is Lebesgue measurable. This holds because of the fact that Lebesgue measurable sets form a $\sigma$-algebra.

If moreover the sets $A_n$ obey $A_n \cap A_m = \emptyset$ for $n \neq m$ then an axiom that by definition must hold for all measures tells us that:

$$\mu(\bigcup_{n \in \mathbb{N}} A_n) =\sum_{n \in \mathbb{N}} \mu(A_n)$$

and this does need disjointness.

For normal unions we can merely say:

$$\mu(\bigcup_{n \in \mathbb{N}} A_n) \le \sum_{n \in \mathbb{N}} \mu(A_n)$$

the so-called sub-additive property of measures (which follows from the above fact for pairwise disjoint sets plus some $\sigma$-algebra manipulation).