how prove that $\lim_{n\to{}\infty} {\sum_{(i, j)\in{K_n}}^\infty{a_{ij}}}=\sum_{i=1}^\infty\sum_{j=1}^\infty{a_ {ij}}$

Question

Let $a_{ij}\geq 0$ $(i,j)\in \mathbb{N}^2$ then $$\sum_{j=1}^\infty\sum_{i=1}^\infty{a_{ij}}=\sum_{i=1}^\infty\sum_{j=1}^\infty{a_{ij}}$$

Moreover, given $ K_1 \subseteq K_2 \subseteq \ldots \subseteq \mathbb{N}^2$ such that $\bigcup_{n\in{\mathbb{N}}}^{}{K_n}=\mathbb{N}^2$ then $\displaystyle \lim_{n\to{}\infty} {\sum_{(i, j)\in{K_n}}^\infty{a_{ij}}}=\sum_{i=1}^\infty\sum_{j=1}^\infty{a_ {ij}}$.

I thought of applying the monotone convergence theorem, but not if this will help solve this problem.? Any ideas thanks

Answer 1

Suppose that one iterated sum converges:

$$\sum_{i=1}^\infty \sum_{j=1}^\infty a_{ij}=\lim_{M \to \infty}\sum_{i=1}^M \sum_{j=1}^\infty a_{ij} = A.$$

Then for all $i \in \mathbb{N}$,

$$S_i = \sum_{j=1}^\infty a_{ij} < \infty.$$

Since $a_{ij} \leqslant S_i$, it follows by the comparison test that

$$T_j =\sum_{i=1}^\infty a_{ij} \leqslant \sum_{i=1}^\infty S_i = A,$$

and, since we can interchange a finite sum with an infinite sum,

$$\sum_{j=1}^NT_j = \sum_{j=1}^N \sum_{i=1}^\infty a_{ij}=\sum_{i=1}^\infty \sum_{j=1}^N a_{ij}\leqslant \sum_{i=1}^\infty S_i = A.$$

The sequence of partial sums on the LHS is monotonically increasing with upper bound $A$ and , hence, convergent, with

$$A' =\sum_{j=1}^\infty \sum_{i=1}^\infty a_{ij} = \lim_{N \to \infty} \sum_{j=1}^N T_j \leqslant A.$$

Repeating the argument with the order of summation over $i$ and $j$ switched we find $A \leqslant A'$.

Therefore,

$$\sum_{i=1}^\infty \sum_{j=1}^\infty a_{ij} = \sum_{j=1}^\infty \sum_{i=1}^\infty a_{ij}.$$

For the second part, let $k_n = \max (\max \{i:(i,j) \in K_n\},\max \{j:(i,j) \in K_n\})$. Then

$${\sum_{(i, j)\in{K_n}}{a_{ij}}}\leqslant \sum_{i=1}^{k_n} \sum_{j=1}^{k_n} a_{ij}\leqslant A.$$

Since $a_{ij} \geqslant 0$ and $K_n \subset K_{n+1}$, the sequence of sums over $K_n$ is monotonically increasing and bounded above by $A$.

Hence,

$$\lim_{n\to{}\infty} {\sum_{(i, j)\in{K_n}}{a_{ij}}}\leqslant\sum_{i=1}^\infty\sum_{j=1}^\infty{a_ {ij}}.$$

To prove equality -- for any $\epsilon > 0$ there exists $P,Q \in \mathbb{N}$ such that

$$A - \epsilon < \sum_{i=1}^P\sum_{j=1}^Q{a_ {ij}} \leqslant A.$$

The nested sequence of sets $K_n$ has $\bigcup_{n\in{\mathbb{N}}}^{}{K_n}=\mathbb{N}^2$. Hence there exists $N \in \mathbb{N}$ such that $\{1,2,\ldots,P\} \times \{1,2,\ldots,Q\} \subset K_n$ for all $n \geqslant N$ and

$$A - \epsilon < \sum_{i=1}^P\sum_{j=1}^Q{a_ {ij}} \leqslant {\sum_{(i, j)\in{K_n}}{a_{ij}}} \leqslant A.$$