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This is from "Measures, Integrals and Martingales" by R.L. Schilling, page 28, exercise 4.6ii.

Prove that for any double sequence $\beta_{ij}, i, j \in \mathbb{N}$ of real numbers, we have

$$ \sup_{i\in\mathbb{N}}\sup_{j\in\mathbb{N}}\beta_{i j} = \sup_{j\in\mathbb{N}}\sup_{i\in\mathbb{N}}\beta_{i j} $$

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In the solution manual to the textbook, we are given half of the proof. I couldn't figure out the other half, but then I did, so I decided to answer my own question:

Note that $\beta_{mn} \leq \sup_{j\in\mathbb{N}}\sup_{i\in\mathbb{N}}\beta_{i j} $ for all choices of $m,n \in \mathbb{N}$. The right hand side of the inequality is independent of $n$ so we can take the supremum over all $n$ i.e.

$$\sup_{n\in\mathbb{N}} \beta_{mn} \leq \sup_{j\in\mathbb{N}}\sup_{i\in\mathbb{N}}\beta_{i j}, \forall m \in \mathbb{N}$$

With the same argument, we can then take the supremum over all $m \in \mathbb{N}$ to get

$$\sup_{m\in\mathbb{N}} \sup_{n\in\mathbb{N}} \beta_{mn} \leq \sup_{j\in\mathbb{N}}\sup_{i\in\mathbb{N}}\beta_{i j}$$

Now, since the names of the indices on both sides of the equation are irrelevant, we can rewrite this as

$$\sup_{i\in\mathbb{N}} \sup_{j\in\mathbb{N}} \beta_{ij} \leq \sup_{j\in\mathbb{N}}\sup_{i\in\mathbb{N}}\beta_{i j}$$ Let $$A = \sup_{i\in\mathbb{N}} \sup_{j\in\mathbb{N}} \beta_{ij}, B = \sup_{j\in\mathbb{N}}\sup_{i\in\mathbb{N}}\beta_{i j}$$

Then

$$A \leq B$$

Now, for the other part of the proof: start off with $\beta_{mn} \leq \sup_{i\in\mathbb{N}}\sup_{j\in\mathbb{N}}\beta_{i j} $ for all choices of $m,n \in \mathbb{N}$ (i.e. we work with the other order of application of supremums). By the same argument, we can take the supremum over all $m \in \mathbb{N}$ i.e.

$$\sup_{m\in \mathbb{N}}\beta_{mn} \leq \sup_{i\in\mathbb{N}}\sup_{j\in\mathbb{N}}\beta_{i j}, \forall n \in \mathbb{N} $$ And then we take the supremum over all $n \in \mathbb{N}$ $$\sup_{n \in \mathbb{N}}\sup_{m\in \mathbb{N}}\beta_{mn} \leq \sup_{i\in\mathbb{N}}\sup_{j\in\mathbb{N}}\beta_{i j}$$ We again rename the indices on the left hand side to obtain $$\sup_{j \in \mathbb{N}}\sup_{i\in \mathbb{N}}\beta_{ij} \leq \sup_{i\in\mathbb{N}}\sup_{j\in\mathbb{N}}\beta_{i j}$$ Which is $$ B \leq A$$ Therefore $A = B$

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