I know that $L^p([0,1])$ is not isometrically isomorphic to $l^p(\mathbb{N})$ when $p\neq 2$? But, there is an isometric copy of $l^p(\mathbb{N})$ inside $L^p([0,1])$. My question is that whether $L^p([0,1])$ can be written as an infinite direct/tensor product of $l^p(\mathbb{N})$ or $L^p([0,1])$ can be built up from countably infinite copies of $l^p(\mathbb{N})$?

Since $[0,1]$ is measurably isomorphic to $2^\mathbb{N}$ with product measure, I am feeling that $L^p([0,1])$ is isometrically isomorphic to infinite tensor product of $ \mathbb{C}^2$ with $l^p$-norm. Is this statement correct?