Euclidean L² product unitary: L²(R^a)⊗̂L²(R^b) ≅ L²(R^{a+b})

[Isometry on finite sums] For $f\in L^2({\lambda }_a)$ and $g\in L^2({\lambda }_b)$ choose measurable representatives. By [F2], replacing them by Borel representatives changes the product only on null strips in the completed product, and the product class is well defined. For finite sums $z={\sum }_{i=1}^n{\alpha }_i{f}_i\otimes g_i$ and $z^{\mathrm{\prime }}={\sum }_{j=1}^m{\beta }_j{f}_j^{\mathrm{\prime }}\otimes g_j^{\mathrm{\prime }}$, each $f_i\overline{f_j^{\mathrm{\prime }}}\in L^1({\lambda }_a)$ and $g_i\overline{g_j^{\mathrm{\prime }}}\in L^1({\lambda }_b)$ by [F1], so the product is integrable in the completed product by Fubini [F2], and ${\int }_{{\mathbb{R}}^{a+b}}{f}_i(x)\overline{f_j^{\mathrm{\prime }}(x)}{g}_i(y)\overline{g_j^{\mathrm{\prime }}(y)}d{\lambda }_{a+b}(x,y)=⟨f_i,f_j^{\mathrm{\prime }}⟩⟨g_i,g_j^{\mathrm{\prime }}⟩\mathrm{.}$ The map $(f,g)\mapsto [f(x)g(y)]$ is bilinear, hence descends through the defining tensor relations to a linear map ${\mathrm{\Theta }}_0$ on the algebraic tensor product. Summing with coefficients ${\alpha }_i\overline{{\beta }_j}$ gives $⟨{\mathrm{\Theta }}_{0}z,{\mathrm{\Theta }}_0{z}^{\mathrm{\prime }}⟩=⟨z,z^{\mathrm{\prime }}⟩$ by [F1]. Taking $z^{\mathrm{\prime }}=z$ gives ${\mathrm{\Theta }}_{0}z=0\Rightarrow z=0$, so ${\mathrm{\Theta }}_0$ is well defined and isometric on the dense algebraic tensor product.

[Finite-positive-measure uniqueness in the first variable] Let $h\in L^2({\mathbb{R}}^{a+b})$ be orthogonal to every product of bounded-box indicators, so ${\int }_{B\times C}h=0$ for all boxes $B,C$. For each bounded rational half-open box $B\subseteq {\mathbb{R}}^b$, define $h_B(x):={\int }_{B}h(x,y)dy$ for a.e. $x$. Cauchy-Schwarz gives $\mathrm{\mid }{h}_B(x){\mathrm{\mid }}^2\le \mathrm{\mid }B\mathrm{\mid }{\int }_B\mathrm{\mid }h(x,y){\mathrm{\mid }}^{2}dy$, and Tonelli [F2] shows $h_B\in L^2({\mathbb{R}}^a)$. Fubini [F2] gives ${\int }_C{h}_B={\int }_{C\times B}h=0$ for every rational half-open box $C\subseteq {\mathbb{R}}^a$. Fix a bounded rational half-open box $K\subseteq {\mathbb{R}}^a$ and write $u=\mathrm{Re}{h}_B$, $v=\mathrm{Im}{h}_B$. By [F5], choose Borel representatives $\tilde{u},\tilde{v}$ equal a.e. to $u,v$. The family ${\mathcal{P}}_K:=\{K\cap C:C\text{ is a rational half-open box in }{\mathbb{R}}^a\}$ is a $\pi$-system containing $K$ and generating the relative Borel $\sigma$-algebra of $K$ by [F3]. The positive and negative parts ${\tilde{u}}_{+},{\tilde{u}}_{-},{\tilde{v}}_{+},{\tilde{v}}_{-}$ are integrable on $K$ because $K$ has finite measure and $h_B\in L^2$. The finite positive Borel measures $A\mapsto {\int }_A{\tilde{u}}_{+}$ and $A\mapsto {\int }_A{\tilde{u}}_{-}$ agree on ${\mathcal{P}}_K$, including the total mass at $K$, since ${\int }_{A}u=0$ there; likewise the measures from ${\tilde{v}}_{+}$ and ${\tilde{v}}_{-}$ agree. By [F3] these equalities hold for every Borel $A\subseteq K$. Taking the Borel sets where $\tilde{u}>0$, $\tilde{u}<0$, $\tilde{v}>0$ and $\tilde{v}<0$ shows $u=v=0$ a.e. on $K$. The countable rational half-open boxes $K$ cover ${\mathbb{R}}^a$, so $h_B=0$ a.e. on ${\mathbb{R}}^a$ for each rational half-open $B$.

[Zero orthogonal complement] There are countably many rational half-open boxes $B\subseteq {\mathbb{R}}^b$, and step 1.2 gives $h_B=0$ a.e. for each; completed-product Fubini [F2] also gives $h(x,\cdot )\in L^2({\mathbb{R}}^b)$ for a.e. $x$. Intersecting these countably many full-measure sets, for a.e. $x$ the section is in $L^2$ and ${\int }_{B}h(x,y)dy=0$ for every rational half-open $B$. Fix such an $x$. On each bounded rational half-open box $K\subseteq {\mathbb{R}}^b$, choose Borel representatives of the real and imaginary parts of $h(x,\cdot )$ by [F5]. The positive and negative parts of these representatives define finite positive Borel measures on $K$; as in step 1.2 they agree on the $\pi$-system of rational half-open subboxes, including $K$, so [F3] makes the positive and negative measures equal on every Borel subset of $K$. Their positive and negative level sets then show the real and imaginary parts vanish a.e. on $K$. The countably many $K$ cover ${\mathbb{R}}^b$, hence $h(x,\cdot )=0$ a.e. for a.e. $x$, and Tonelli [F2] gives $h=0$ a.e. Thus $\mathrm{Ran}{\mathrm{\Theta }}_0^{\perp }=\{0\}$; since $\mathrm{Ran}{\mathrm{\Theta }}_0$ is a linear subspace, [F6] gives $\overline{\mathrm{Ran}{\mathrm{\Theta }}_0}=\mathrm{Ran}{\mathrm{\Theta }}_0^{\perp \perp }=L^2({\lambda }_{a+b})$.

[Extension and surjectivity] By [F4], whose Banach-space hypothesis holds by Riesz–Fischer, the isometry ${\mathrm{\Theta }}_0$ on the dense algebraic tensor product extends uniquely to a linear isometry $\mathrm{\Theta }$ of the completion. Its range is closed (isometries have closed range: if $\mathrm{\Theta }{z}_n\to w$ then $(z_n)$ is Cauchy, so $z_n\to z$ and $w=\mathrm{\Theta }z$) and dense by step 2.1, hence is all of $L^2({\lambda }_{a+b})$. Thus $\mathrm{\Theta }$ is a surjective linear isometry, i.e. unitary, with the stated action on elementary tensors.