If $\lambda=$ measure of a set and all $G_k$'s are open sets, then : $\lambda ( \cup_{k=1}^{\infty} G_k ) \le \sum _{k=1}^{\infty}\lambda ( G_k)$ I just started reading the book Lebesgue Integration on Euclidean Spaces by Frank jones, in which the author gives a result and it's proof as : the If $\lambda$ denotes the measure of a set and all $G_k$'s are open sets, then : $$\lambda ( \bigcup_{k=1}^{\infty} G_k ) \le  \sum _{k=1}^{\infty}\lambda ( G_k)$$
A special polygon is any polygon which can be divided further into finite number of rectangles whose sides are parallel to the co-ordinate axes.
Proof:


Query:

The red boxed area says that it's possible to define $P_k$ to be the union of all $I_j's$ such that $I_j \subset G_k$ and $I_j \nsubseteq G_1,G_2, \cdots G_{k-1}~~~~~~ \dots \dots (A)$

Suppose $I_1,I_2 \in G_3$ and no other $G_k:k \in \mathbb N$ contains $I_2$.
Suppose $I_1, I_3 \in G_4$ and no other $G_m:m \in \mathbb N$ contains $I_3$.
Hence, we can't ignore $G_3$ nor $G_4$ because even though they contain duplicate of $I_1$ , they also contain $I_2$ and $I_3$ respectively which no other $G_k$ contains. Keeping in mind requirement $(A)$, is this not a contradiction with the requirement (A)?
Could someone be please kind enough to tell me where I might be going wrong or if the proof might have a glitch?Thank you for reading!

I don't see a problem.  In the example you give, assuming $P = I_1 \cup I_2 \cup I_3$, we would take $P_1 = P_2 = \emptyset$, $P_3 = I_1 \cup I_2$, and $P_4 = I_3$, and then $P_5 = P_6 = \dots = \emptyset$.
Maybe you are misinterpreting the sentence "Since each $I_j$ is contained in one of $G_1, G_2, \dots$."  What Jones means is "Each $I_j$ is contained in at least one of $G_1, G_2, \dots$" - there is no problem if some $I_j$ is contained in several of the $G_k$, as long as it doesn't get missed altogether.  This is just to explain the assertion that $P = \bigcup_k P_k$.
