Definition 1. Let the set
$$S := \{S \subset \mathbb{R}^2 ; R = \{(x,y) \in \mathbb{R}^2 ; a \leq x \leq b \ \text{and} \ c \leq y \leq d \}\}.$$
Definition 2. We define that $m(\emptyset) = 0$ and $m(R) = (d - c)(b - a)$, for every $R \in S \ne \emptyset$.
Definition 3. We say that a subset $A \subset \mathbb{R}^2$ is elementary if the set is a finite union of disjoint elements of $S$.
Definition 4. Let $A = \cup_{i = 1}^n S_i$ be a elementary set. Then $m'(A) = \sum_{i = 1}^n m(S_i)$.
I'm studying for a summer of Measure Theory and I came across this Theorem:
Let $A_1, A_2, ..., A_n$ be a sequence of disjoint elementary sets. If $A = \cup A_i$ is elementary, then $m'(A) = \sum_{i = 1}^\infty m'(A_i)$
My proof:
- $m'(A) \geq \sum_{i = 1}^\infty m'(A_i)$.
For every $n \in \mathbb{N}$, take
$$B_n = A - \cup_{i = 1}^n A_n.$$
Then $B_n$ is elementary and $A = B_n \cup A_1 \cup ... \cup A_n$. Note that
$$m'(A) = m'(B_n) + \sum_{i = 1}^n m'(A_i) \geq \sum_{i = 1}^n m'(A_i), $$
for all $n \in \mathbb{N}$. Therefore
$$m'(A) \geq \sum_{i = 1}^\infty m'(A_i).$$
My doubt is: How do I prove the opposite?
The professor suggested that we prove the following Lemma:
Lemma. Let $A, A_1, A_2, ..., A_n, ...$ be a elementary sets such that $A \subseteq \cup_{i = 1}^\infty A_i$. Then $m'(A) \leq \sum_{i = 1}^\infty m'(A_i).$
I would be grateful if you could help me prove the opposite inequality with any suggestions or indicating where I can find this proof. Thanks.