Grothendieck's lemma in $L^p$ spaces

So I am currently working on the proof of Grothendieck's Lemma :

Let S $$\subset L^{\infty}(X)$$, of finite measure, be a closed vector subspace of $$L^p$$ for a certain p such that $$S \subset L^{\infty}$$

We want to show that S is finite dimensional.

Suppose that the embedding $$L^2 \subset L^{\infty}$$ is continuous.

We give an orthonormal family of vectors of S.

$$\forall c = (c_1, ..., c_n) \in \mathbb{Q}^n$$ (countable and dense in $$\mathbb{R}^n$$ ) $$\exists$$ $$X_c \subset X$$ of full measure, such that

$$\forall x \in X_C$$ , $$| \sum_i c_i f_i (x) | \leq M || \sum_i c_i f_i ||_{\infty} \leq M || \sum_i c_i f_i ||_{2}$$

We also have $$\forall c \in \mathbb{Q}^n \: \: \forall x \in \cup_{c \in \mathbb{Q}^n } X_c$$ , $$| \sum_i c_i f_i (x) | \leq M || \sum_i c_i f_i ||_{\infty} \leq M || \sum_i c_i f_i ||_{2}$$

So as $$\mathbb{Q}^n$$ is countable and dense in $$\mathbb{R}^n$$ we get the same result for all $$c \in \mathbb{R}^n$$

In particular for $$c_i = f_i (x)$$ we get $$\sum_i (f_i (x))^2 \leq M \sqrt{\sum_i f_i ^2}$$ so $$\sum_i (f_i (x))^2 \leq M^2$$

By integration on $$\cup_{c \in \mathbb{Q}^n } X_c$$ we deduce that : $$n \leq M^2$$

• my question : Wy did we take $$\mathbb{Q}^n$$ countable and dense ?

Apparently as $$X_c$$ depends on c we would get an issue trying that integration.

We also need to use the definition of $$|| . ||_{\infty}$$ I suppose.

But is there something to do with a Quantifier inversion? Why specificaly a countable set?

The simple reason for varying $$c$$ over a countable set is that an uncountable union of sets of measure $$0$$ need not have measure $$0$$. [ Like the singleton sets in $$\mathbb R$$]. In the line next to the one where $$X_c$$ is introduced we want the inequality to hold for $$x$$ outside one null set $$X_0$$ for all $$c$$. But the union of the sets $$X_c^{c}$$ need not have measure $$0$$ if we vary $$c$$ over an uncountable set. So we first vary $$c$$ over $$\mathbb Q^{n}$$ and then observe that if the inequality we get holds for $$c$$ in a dense set of $$c$$'s it holds for all $$c$$ by a continuity argument.