Invariant square integrable kernel produces a compact intertwiner

The preimage under $T\times T$ of $E\times F$ is $T^{-1}E\times T^{-1}F$, of the same measure. These rectangles generate the product sigma-algebra and include the whole probability square, so F6 applies. Preimages of completed null subsets lie in the preimages of product-measurable null covers, hence are measurable and null in the completion. Thus $T\times T$ preserves the completed product. Its Koopman operator $W$ and the factor operator $U$ are isometries. Obtain $V=U^{\ast }$ and $VU=I$ from F3, under AC.

For a rectangle tensor $l(x,y)=a(x)b(y)$, the adjoint identity and its conjugate give $\int b(Ty)f(y)d\mu (y)=⟨f,U\bar{b}⟩=⟨Vf,\bar{b}⟩=\int b(y)Vf(y)d\mu (y)$. Multiplying by $a(Tx)$ yields $K_{Wl}=U{K}_{l}V$. Finite linear combinations obey the same identity. For any $l\in L^2$, approximate by such combinations in kernel norm; isometry of $W$ and F1 make both sides converge in operator norm. Hence the identity holds for every kernel.

Since $Wk=k$, step 2.1 gives $K=UKV$, and multiplying on the right by $U$ gives $KU=UK$. The swapped conjugate kernel obeys $W{k}^{\ast }=k^{\ast }$ too: F2 makes conjugate transpose a well-defined operation on completed $L^2$ kernel classes, algebraically $W(k^{\ast })=(Wk{)}^{\ast }$, and $Wk=k$. Applying the same identity to $k^{\ast }$ yields $K^{\ast }U=U{K}^{\ast }$. Compactness of both operators comes from F1–F2.

Vanishing $y$-marginal gives $K\mathbf{1}=0$. Vanishing $x$-marginal gives $K^{\ast }\mathbf{1}=0$ after conjugation. For $f\in H_0$, $⟨Kf,\mathbf{1}⟩=⟨f,K^{\ast }\mathbf{1}⟩=0$, and similarly for $K^{\ast }$ using its double adjoint. Thus both preserve $H_0$. If $k\mathrm{\ne }0$, F1 gives $K\mathrm{\ne }0$. Every $f$ splits as $(\int f)\mathbf{1}+f_0$ with $f_0\in H_0$; since $K\mathbf{1}=0$, a vector with $Kf\mathrm{\ne }0$ supplies $K{f}_0\mathrm{\ne }0$. Therefore the restriction is nonzero. All marginal equalities are a.e. equalities of integrable sections, justified by F7.