This is part of a problem I am working on from an old prelim exam in analysis, to prepare for my own prelim. I asked about another part of this question here: Norm of an Operator on $L^2(\mu)$
We are given that $(X,\Omega,\mu)$ is a $\sigma-$finite measure space. We define $A: L^2(\mu)\to L^2(\mu)$ by $$ (Af)(x) = \int_X K(x,y)f(y)d\mu(y) $$ where $K\in L^2(\mu\times\mu)$.
We are asked to prove that the correspondence $K\to A$ is one-to-one. I am very surprised this statement is true---can't $K_1$ and $K_2$ differ merely on a set of $\mu-$measure $0$ and thus still generate the same $A$?
Assuming the statement is true, we can let $K_1,K_2\in L^2(\mu\times\mu)$ with $$ \int_X K_1(x,y)f(y)d\mu(y) = \int_X K_2(x,y)f(y)d\mu(y)\quad\forall x $$ From here, how should I go about trying to show that $K_1=K_2$?
If $K_1$ and $K_2$ differ on a set of $\mu\times\mu$-measure zero then they define the same element of $L^2(\mu\times\mu)$.
To show the correspondence is one-to-one you need to show that if $A=0$ then $K=0$ ($\mu\times\mu$-almost everywhere).
First, if $E$ is a measurable subset of $X\times X$ and $E$ has finite measure then $\chi_E$ can be approximated in $L^2$ by a linear combination of characteristic functions of measurable rectangles $A\times B$; this follows from the definition of the product measure.
So linear combinations of functions of the form $f(x)g(y)$ are dense in $L^2(\mu\times\mu)$. It follows that if $e_n(x)$ is an orthonormal basis for $L^2(\mu)$ then $e_n(x)e_m(y)$ is an orthonormal basis for $L^2(\mu\times\mu)$.
Suppose $K\ne0$. Let $$K=\sum a_{n,m} e_{n}(x)e_{m}(y).$$ It's now easy to show that there exists $m$ with $Ae_m\ne 0$.