Is this definition of countable subadditivity correct?

Question

From Wolfram:

However, I believe that the countable the sets in $\{E_k\}_{k=1}^n$ do not have to be disjoint, at least not for outer measure (the precursor to Lebesgue measure). Am I correct?

Answer 1

Whether you require that the sets are disjoint doesn't really matter for measures. Say that $\mu$ is disjointly-subadditive to mean that $\mu$ has the subadditivity property for disjoint families, and subadditive if we don't have that restriction. Then they are equivalent, assuming that $\mu$ is a nonnegative function.

$(\implies)$ If $\mu$ is disjointly-subadditive and $\{E_k\}$ is a countable collection of sets, we can inductively choose disjoint sets $A_k \subseteq E_k$ which are disjoint and $\bigcup E_k = \bigcup A_k$. Then $$\mu\left(\bigcup_k E_k\right) = \mu\left(\bigcup_k A_k\right) \le \sum_k \mu(A_k) \le \sum_k \mu(E_k)$$ by monotonicity.

$(\impliedby)$ This is obvious, since every disjoint collection is a collection.

It's left for the reader to verify that in either definition, $\mu$ must take only nonnegative values.

Answer 2

You're right that sets $\{ E_k \}$ need not be disjoint.

The above definition for disjoint sets implies subadditivity for non-disjoint sets as well.

Any sequence $\{ E_k \}$ can be written as a disjoint sequence of sets

$$E_1, E_2 - E_1, E_3 - E_2 - E_1$$

So that you can apply subaddivity to get the above inequality. Then apply monotinicity of measures to get the result for the original $E_k$, since $E_k - E_{k-1} - \dots -E_1 \subset E_k$.