Clarification In Proof of Compactness of the Integral Operator with L^2 kernel

Question

I'm doing some self studying out of Axler's Measure and Integration, and the following proof has me a bit confused. Here's the theorem and proof.

I'm specifically confused in the highlighted regions.

1. In the first highlighted section, is Axler saying that $K(x,y) = \lim_{m \rightarrow \infty} K_m(x,y)$ for $K_m(x,y) = \sum_{i=1}^{j_m} g_i(x)h_i(y)$ for $g_i,h_i \in L^2(\mu)$, or is he saying that $K(x,y) = \lim_{m \rightarrow \infty} K_m(x,y)$ for $K_m \in L^2(\mu \times \mu)$ compact?

If we assume he means the first, why is it in the next highlighted region can we assume that $K(x,y) = g(x)h(y)$? Wouldn't $K$ be of the form $K(x,y) = \lim_{m \rightarrow \infty} \sum_{i=1}^{j_m} g_i(x)h_i(y)$?

2. There are three major theorems applied in succession, yet I can only see Fubini's theorem being applied in the equality above the underlined region. Can someone point out in detail how all three theorems are being applied at each step?

3. We showed that if $F$ is orthogonal to every $K \in L^2(\mu \times \mu)$ previously considered, then $F$ must be the $0$ function in $L^2(\mu \times \mu)$ (bottom highlighted region). So then the orthogonal compliment of the subspace $A = \{K \in L^2(\mu \times \mu) \: | \: K(x,y) = \lim_{m \rightarrow \infty} K_m(x,y), \: K_m(x,y) = \sum_{i=1}^{j_m} g_i(x)h_i(y)\}$ must be contained in $\{0\}$, implying that $A^{\perp} = \{0\}$. Hence, $(A^{\perp})^{\perp} = A = L^2(\mu \times \mu)$? If so, why is $A$ nessiarily closed? That doesn't seem immediately obvious to me.

Any help would be appreciated. Thank you.

Answer 1

1. The first that you said. You basically want to approximate the operator with finite rank operators to prove that it is compact, so you need the approximating operators to be of that form. I don’t understand what you wrote between 1 and 2.
2. Mhh, in general to apply Fubini you need to know that the integrand is in $L^1$ of both variables. And to prove that, after putting the absolute value on the integrand, one often (both in theory and in exercises/applications) has to use Tonelli to say that the $L^1$ norm (as a function of both variables) is finite. The thing you might want to prove in this case, though, is actually that $g(x)h(y)$ is in $L^2(\mu\times\mu)$, since you already know that F is in $L^2(\mu\times\mu)$ and you can use Holder to estimate the $L^1$ norm of the product $g(x)h(y)F(x,y)$. To prove that $g(x)h(y)$ is in $L^2(\mu\times\mu)$ you just have to use Fubini (and Tonelli as I said before). Not entirely sure, as I don’t know what the author has already proved before this theorem, but it should be something like that.
3. A is defined as the closure of the span of all the functions of the form 10.71, so it is closed. Just to state it clearly, the orthogonal of the ortogonal of a vector subspace $A$ of a Hilbert space $H$ is $A$ itself if and only if $A$ is closed.in $H$.