Justification for the swap of summations? Im taking a look at the book An introduction to measure and integration of Rana and in the first pages I find the following proof about the length function $\lambda$ on $\Bbb R$:

Property 8: let $I\in\mathcal I$ and $I\subseteq\bigcup_{n=1}^\infty I_n$ where $I_n\in \mathcal I$. Then
$$\lambda(I)\le\sum_{n=1}^\infty \lambda(I_n)$$
This property is called the countable subadditivity of $\lambda$

Here $I$ means an interval in the extended real line and $\mathcal I$ is the set of all intervals of the extended real line. Then the book proved the case for $I$ infinite, that is, $-\infty$ or $+\infty$ is one of the extremes values of $I$:

[...] The case that we need to consider is when $I$ is infinite and all $I_n$ are finite. In this case we write $$I=\bigcup_{k=-\infty}^{+\infty}(I\cap[k,k+1))$$
and notice that $I\cap[k,k+1)\subseteq\bigcup_{n=1}^\infty (I\cap[k,k+1))$. Thus $$\begin{align}\lambda(I)&=\sum_{k=-\infty}^\infty\lambda(I\cap[k,k+1))\\&\le\sum_{k=-\infty}^\infty\sum_{n=1}^\infty\lambda(I_n\cap[k,k+1))\\&=\sum_{n=1}^\infty\sum_{k=-\infty}^\infty\lambda(I_n\cap[k,k+1))\\&=\sum_{n=1}^\infty\lambda(I_n)\end{align}$$

The book doesnt explain why we can change the order of summation. The unique rule that I know to change freely the order of a double series is if
$$\sup_{n\in\Bbb N}\sum_{j,k=1}^n|a_{j,k}|<\infty$$
but clearly this is not the case in the above proof. Can someone show me why the discussed identity is true?
 A: Let it be that for $i,j\in\mathbb N$ every $a_{i,j}$ is non-negative element of $\mathbb R$.
It is evident that: $$\sum_{i=1}^n\sum_{j=1}^ma_{i,j}=\sum_{j=1}^m\sum_{i=1}^na_{i,j}$$ so it is enough to prove that:$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}=\sup\left(\{\sum_{i=1}^n\sum_{j=1}^ma_{i,j}\mid n,m\in\mathbb N\}\right)$$
Also it is evident that $\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}$ is an upper bound of $\{\sum_{i=1}^n\sum_{j=1}^ma_{i,j}\mid n,m\in\mathbb N\}$ so it remains to prove $s$ is not an upper bound of this set whenever $s<\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}$.
If indeed $s<\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i,j}$ then we can find some $n\in\mathbb N$ such that $s<\sum_{i=1}^{n}\sum_{j=1}^{\infty}a_{i,j}$.
Then we can also find a positive $\epsilon$ with $s+n\epsilon<\sum_{i=1}^{n}\sum_{j=1}^{\infty}a_{i,j}$.
For each $i\in\{1,\dots,n\}$ we can find some $m_i\in\mathbb N$ with $\sum_{j=1}^{\infty}a_{i,j}<\sum_{j=1}^{m_i}a_{i,j}+\epsilon$. 
This leads to: $$s+n\epsilon<\sum_{i=1}^{n}\sum_{j=1}^{\infty}a_{i,j}<\sum_{i=1}^{n}\sum_{j=1}^{m_i}a_{i,j}+n\epsilon$$
Then: $$s<\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i,j}$$ where $m:=\max(m_1,\dots,m_n)$.
This shows that $s$ is not an upperbound of the set, and we are ready.
