I'm trying to show that a certain Hilbert-Schmidt operator is compact following some exercises in Rudin's Functional Analysis (exercise 15 on page 112):
If $X, \mu$ is a finite measure space and $K \in L^2 (X \times X)$, define $T: L^2 (X) \rightarrow L^2 (X)$ by
$$T(f)(s) = \int_X K(s,t)f(t) d\mu (t)$$.
I think I'm supposed to show $T$ is compact following this outline:
Approximate $K$ (in some sense) by a sequence $$K_n (s,t) = \sum _{i \leq n} a^n_i (s) b^n_i (t)$$
Where $a^n_i, b^n_i \in L^2 (X)$. Then the operators $$T_n (f) (s) := \int_X K_n(s,t)f(t) d\mu (t) $$ will hopefully converge to $T$ in the operator norm. Then the fact that every $T_n$ has finite dimensional range will imply that $T$ is compact.
The issue is that I don't know how to find $a^n_i$ and $b^n_i$. Thanks for reading my question.
Best Answer
It is in the very definition of product measure space that elementary sets (i.e. set of the form $A\times B$ where $A$ and $B$ are measurable subsets of $X$) generate $\sigma$-algebra of $X\times X$. As the consequence any measurable $C\subset X\times X$ can be "approximted" by finite disjoint unions of elementary sets. Hence so does any $K\in L_2(X\times X,\mu\times \mu)$.
P.S.
There is no finiteness issues here because the original space $X$ is of finite measure.