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Equivalently, if $\Re p>0$, we can drop the boundedness condition on $f$ and finiteness condition on $μ$ and instead require that the measure $|f|^{1/\Re p}μ$ is finite. This produces the same $\L^p$-space because $$(f,μ)=(\arg f\cdot (|f|^{1/\Re p})^{-\Im p}(|f|^{1/\Re p})^p,μ)\sim(\arg f\cdot (|f|^{1/\Re p})^{-\Im p},|f|^{1/\Re p}μ),$$ and in the last pair, the first component has absolute value at most 1 (hence is bounded), whereas $|f|^{1/\Re p}μ$ is a finite measure. The $\L^p$-norm in this description is simply the $\Re p$-th power of the integral of $|f|^{1/\Re p}μ$.

Equivalently, if $\Re p>0$, we can drop the boundedness condition on $f$ and finiteness condition on $μ$ and instead require that the measure $|f|^{1/\Re p}μ$ is finite. This construction produces the same $\L^p$-space because $$(f,μ)=(\arg f\cdot (|f|^{1/\Re p})^{-\Im p}(|f|^{1/\Re p})^p,μ)\sim(\arg f\cdot (|f|^{1/\Re p})^{-\Im p},|f|^{1/\Re p}μ),$$ and in the last pair, the first component has absolute value at most 1 (hence is bounded), whereas $|f|^{1/\Re p}μ$ is a finite measure. Here and below we write $p=\Re p+\Im p$ for the real and purely imaginary parts of $p$. The $\L^p$-norm in this description is simply the $\Re p$-th power of the integral of $|f|^{1/\Re p}μ$.

Equivalently, we can drop the boundedness condition on  $f$ and finiteness condition on  $μ$ and instead require that the measure $|f|^{1/\Re p}μ$ is finite. This produces the same $\L^p$-space. The $\L^p$-norm in this description is simply the $\Re p$-th power of the integral of $|f|^{1/\Re p}μ$.

To see that the resulting $\L^p$-space is isomorphic to the traditional $\def\LL{{\rm L}}\LL^p$-space $$\LL^p(X,M,μ)=\left\{f\biggm| \int|f|^{1/p} dμ<∞\right\},$$ observe that for any faithful finite measure $μ$ the map $$f↦(f,μ)$$ yields an isomorphism $$\LL^p(X,M,μ)→\L^p(X,M,N),$$ where the right side does not depend on the choice of $μ$. This can be easily extended to nonfinite measures $μ$.

Equivalently, if $\Re p>0$, we can drop the boundedness condition on  $f$ and finiteness condition on  $μ$ and instead require that the measure $|f|^{1/\Re p}μ$ is finite. This produces the same $\L^p$-space because $$(f,μ)=(\arg f\cdot (|f|^{1/\Re p})^{-\Im p}(|f|^{1/\Re p})^p,μ)\sim(\arg f\cdot (|f|^{1/\Re p})^{-\Im p},|f|^{1/\Re p}μ),$$ and in the last pair, the first component has absolute value at most 1 (hence is bounded), whereas $|f|^{1/\Re p}μ$ is a finite measure. The $\L^p$-norm in this description is simply the $\Re p$-th power of the integral of $|f|^{1/\Re p}μ$.

To see that the resulting $\L^p$-space is isomorphic to the traditional $\def\LL{{\rm L}}\LL^p$-space $$\LL^p(X,M,μ)=\left\{f\biggm| \int|f|^{1/p} dμ<∞\right\},$$ observe that for any faithful measure $μ$ the map $$f↦(f,μ)$$ yields an isomorphism $$\LL^p(X,M,μ)→\L^p(X,M,N),$$ where the right side does not depend on the choice of $μ$.

To see that the resulting $\L^p$-space is isomorphic to the traditional $\def\LL{{\rm L}}\LL^p$-space $$\LL^p(X,M,μ)=\left\{f\biggm| \int|f|^{1/p} dμ<∞\right\},$$ observe that for any faithful finite measure $μ$ the map $$f↦(f,μ)$$ yields an isomorphism $$\LL^p(X,M,μ)→\L^p(X,M,N),$$ where the right side does not depend on the choice of $μ$. This can be easily extended to nonfinite measures $μ$.

The Hölder inequality yields a canonical multiplication map $$\L^p(X,M,N)⊗\L^q(X,M,N)→\L^{p+q}(X,M,N),$$ and the induced map $$\L^p(X,M,N)⊗_{\L^0(X,M,N)}\L^q(X,M,N)→\L^{p+q}(X,M,N)$$ is an isomorphism, where $⊗$ denotes the usual algebraic tensor product, which happens to be automatically complete.

Equivalently, we can drop the boundedness condition on $f$ and finiteness condition on $μ$ and instead require that the measure $|f|^{1/\Re p}μ$ is finite. This produces the same $\L^p$-space. The $\L^p$-norm in this description is simply the $\Re p$-th power of the integral of $|f|^{1/\Re p}μ$.

To see that the resulting $\L^p$-space is isomorphic to the traditional $\def\LL{{\rm L}}\LL^p$-space $$\LL^p(X,M,μ)=\left\{f\biggm| \int|f|^{1/p} dμ<∞\right\},$$ observe that for any faithful finite measure $μ$ the map $$f↦(f,μ)$$ yields an isomorphism $$\LL^p(X,M,μ)→\L^p(X,M,N),$$ where the right side does not depend on the choice of $μ$. This can be easily extended to nonfinite measures $μ$.

The Hölder inequality yields a canonical multiplication map $$\L^p(X,M,N)⊗\L^q(X,M,N)→\L^{p+q}(X,M,N)$$ given by $$((f,μ),(g,μ))↦(fg,μ),$$ where we used the equivalence relation $\sim$ to pick representatives for elements of $\L^p$ and $\L^q$ with the same measure $μ$ as the second component. The induced map $$\L^p(X,M,N)⊗_{\L^0(X,M,N)}\L^q(X,M,N)→\L^{p+q}(X,M,N)$$ is an isomorphism, where $⊗$ denotes the usual algebraic tensor product, which happens to be automatically complete.