Tensor products of sigma-finite L2 spaces

Measurable representatives may be chosen for $f,f^{\mathrm{\prime }},g,g^{\mathrm{\prime }}$. Replacing a representative on a null set changes its product only on a product-null set: cover the other sigma-finite factor by countably many finite-measure sets and apply the rectangle formula [F1] on each. Thus the product rule is well defined on $L^2$ classes. By [F7], $f\overline{f^{\mathrm{\prime }}}$ and $g\overline{g^{\mathrm{\prime }}}$ are integrable; Fubini [F2] gives the product norm $\mathrm{\parallel }f(x)g(y){\mathrm{\parallel }}_2^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2\mathrm{\parallel }g{\mathrm{\parallel }}_2^2$, so each elementary product belongs to the target space. It also gives $⟨f(x)g(y),f^{\mathrm{\prime }}(x)g^{\mathrm{\prime }}(y){⟩}_{L^2(X\times Y)}=⟨f,f^{\mathrm{\prime }}{⟩}_{H_X}⟨g,g^{\mathrm{\prime }}{⟩}_{H_Y}\mathrm{.}$ By sesquilinearity, the rule ${\sum }_j{\alpha }_j{f}_j\otimes g_j\mapsto {\sum }_j{\alpha }_j{f}_j(x)g_j(y)$ preserves inner products on finite sums. If two tensor presentations define the same algebraic tensor, their difference has zero tensor norm by [F5], hence its product function has zero $L^2$ norm; the rule is well-defined on the algebraic tensor product and extends uniquely as an isometry $T$ to the completion.

Choose sigma-finite exhaustions $X_k\uparrow X$ and $Y_k\uparrow Y$ with finite measure, and put $Z_k=X_k\times Y_k$. By the rectangle formula [F1], $(\mu \times \nu )(Z_k)<\mathrm{\infty }$, and $Z_k\uparrow X\times Y$. For fixed $k$, let ${\mathcal{C}}_k$ be the family of product-measurable $E\subseteq Z_k$ such that ${\mathbf{1}}_E$ belongs to the closed range of $T$. It contains $Z_k$, since ${\mathbf{1}}_{Z_k}=T({\mathbf{1}}_{X_k}\otimes {\mathbf{1}}_{Y_k})$. If $E\subseteq F$ are in ${\mathcal{C}}_k$, then ${\mathbf{1}}_{F\setminus E}={\mathbf{1}}_F-{\mathbf{1}}_E$ is in that closed range. If $E_j\uparrow E$ in $Z_k$, then $\mathrm{\parallel }{\mathbf{1}}_E-{\mathbf{1}}_{E_j}{\mathrm{\parallel }}_2^2=(\mu \times \nu )(E\setminus E_j)\to 0$ by [F6], so $E\in {\mathcal{C}}_k$. Thus ${\mathcal{C}}_k$ is a lambda-system on $Z_k$. For every measurable rectangle $A\times B$, its restriction to $Z_k$ is $(A\cap X_k)\times (B\cap Y_k)$, whose indicator is the image of ${\mathbf{1}}_{A\cap X_k}\otimes {\mathbf{1}}_{B\cap Y_k}$; these restricted rectangles are a generating pi-system by [F4]. Dynkin's theorem [F4] gives ${\mathbf{1}}_E\in \overline{\mathrm{ran}T}$ for every product-measurable $E\subseteq Z_k$.

Let $E$ be any completed-product-measurable set with finite product measure. By the completion definition [F1], it differs from a product-measurable set $E_0$ by a subset of a product-null set, so ${\mathbf{1}}_E={\mathbf{1}}_{E_0}$ as $L^2$ classes. Since $Z_k\uparrow X\times Y$, [F6] gives $(\mu \times \nu )(E_0\setminus Z_k)\to 0$, so ${\mathbf{1}}_{E_0}={\lim }_k{\mathbf{1}}_{E_0\cap Z_k}$ in $L^2$; step 2.1 puts each truncated indicator in the closed range of $T$. Every finite-measure simple function for the completed measure is a finite linear combination of indicators of completed-measurable finite-measure sets; replacing each set by its product-measurable representative gives the same $L^2$ class. Such simple functions are dense by [F3]. Consequently the range of $T$ is dense in the full target $L^2$ space. Since the range of an isometry from a complete space is closed, $T$ is onto and therefore unitary. Its uniqueness follows because elementary tensors span a dense subspace.