Is a measure of a union of sets is equal to the sum of measures on individual sets IFF the sets are disjoint?

Question

Background

A (finite) measure is countably additive by definition. This entails that for a collection of disjoint sets the measure on the union of those sets is equal to the sum of the measures on each set. The IE principle allows us to work out the measure on the union when the sets are not disjoint.

Question

For a collection of sets, if I find that

$$\mu \left( \bigcup_{k}^{\infty} E_k \right) = \sum_{k}^{\infty} \mu \left( E_k \right)$$

holds, does that imply that $E_k$ are disjoint? Or can there be some fact about how various non-empty intersections cancelling out in their measure (thinking of the IE principle) that produces the equality?

Answer 1

Let $E_k$ be a (finite or countably infinite) collection of sets. Call $E$ their union. We suppose $E$ has finite measure.
$E$ can be partitioned into non-empty sets $F_j$, where each $F_j$ is composed of all elements that belong to the same list of $E_k$ sets. We suppose that all $F_j$ have non-null measure.

As sets $F_j$ are disjoint, $\mu (E) = \mu \left( \bigcup_j F_j \right) = \sum_{j} \mu \left( F_j \right)$.

In $\sum_{j} \mu \left( F_j \right)$ we sum positive values, so we can reorder the summation.

If $E_k$ are non disjoint, there exists some $F_j$ whose elements belong to more than one $E_k$.

• In $\sum_{j} \mu \left( F_j \right)$, this $F_j$ appears only once.
• In $\sum_{k} \mu \left( E_k \right)$, this $F_j$ appears a number of times equal to the number of $E_k$'s to which its elements belong.

As this $F_j$ has a measure $> 0$, $\sum_{k} \mu \left( E_k \right) \gt \sum_{j} \mu \left( F_j \right) = \mu(E) = \mu \left( \bigcup_k E_k \right)$.

$\\$

Note: the hypothesis "all $F_J$ have non-null measure" is necessary.
The hypothesis "$E_k$ are a finite or countably infinite collection" is perhaps not necessary, but I don't feel like digging into that direction. ;-)