Question about example of non-separable Hilbert space

I have come across the following example of a non-separable Hilbert space:

Example 2.84. Let $I$ be a set, equipped with the discrete topology and the counting measure $\lambda_{\text{ count}}$ defined on the $\sigma$-algebra $\Bbb P(I)$ of all subsets of $I$. Then $$\ell^2(I)=L^2\big(I,\Bbb P(I),\lambda_{\text{ count}}\big)$$ is a Hilbert space, and it comprises all functions $a:I\to\Bbb R$ (or $\Bbb C$) for which the support $$F=\{i\in I : a(i)\ne0\},$$ is finite or countable, and for which $\sum_{i\in I}|a_i|^2=\sum_{i\in F}|a_i|^2\lt\infty$.

Why do I need the discrete topology on $I$? Or more generally: why do I need a topology? If we talk about $L^p$ spaces in general, we only want a measure space and we don't mention a topology because $f \in L^p$ doesn't have to be continuous. Thanks for your help.

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We do not need any topology here – Norbert Aug 2 '12 at 16:19
You are welcome! – Norbert Aug 2 '12 at 16:21
Maybe topology is useful in an other part of the example (if the whole example is displayed, I don't see). A remark: the Hilbert space we define in such a way can be separable (when $I$ is finite or countable), otherwise it isn't. – Davide Giraudo Aug 2 '12 at 16:22
Maybe earlier examples were about $L^p$-spaces with respect to Radon measures on locally compact spaces? The discrete topology is locally compact and $\mathbb P(I)$ is the Borel $\sigma$-algebra, etc. hence compactly supported (=finitely supported) functions are dense. – t.b. Aug 2 '12 at 16:27
I don't see a typo here. It is in the mindset of many people to equip a set with the discrete topology by reflex when it has none and to say that explicitly. That's why you find "Let $G$ be a discrete group" all over the place even when there's no topological group whatsoever in sight. – t.b. Aug 2 '12 at 17:28

You are correct, the topology on $I$ is irrelevant and was probably mentioned by mistake.

(If someone will upvote this answer, the question will stop being bumped.)

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In fact, "discrete topology on $I$ = $\sigma$-algebra of $\mathbb{P}(I)$ of all subsets of $I$".

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The reason why a discrete topology is required for $I$, is to make sure that the $\sigma$-algebra of the Borel sets is the whole power set of $I$, i.e. ${\mathscr P}(I)$.

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