# $L^2$ norm on a product space $]0,1[\times \Omega$

Let ($\Omega, \mathcal A, P)$ be a measure space, and $]0,1[$ equipped with the corresponding Lebesgue $\sigma$-algebra and Lebesgue measure.

We have a function $f:]0,1[ \times \Omega \to \mathbb R$

What does mean $$\Vert f \Vert_{L^2(]0,1[\times \Omega)}$$

And is it the same as (or does this expression make sense) $$\Vert f \Vert_{L^2(\Omega\times ]0,1[)}$$

$$\lVert f\rVert_{L^2(]0,1[\times\Omega)}=\sqrt{\int_{]0,1[\times\Omega} \lvert f\rvert^2\,dx\otimes dP}$$ where $dx\otimes dP$ indicates the product measure on the product $\sigma$-algebra of $]0,1[\times\Omega$. Namely, the product $\sigma$-algebra is the one generated by products of measurable sets. The product measure is the only measure on that $\sigma$-agebra such that $\mu(A\times B)=\mathcal L(A)\times P(B)$ for all $A,B$ measurable in the original spaces (with the notational assumption that $0\cdot M=0$ for all $M$ including $\infty$).
• So $\lVert f\rVert_{L^2(]0,1[\times\Omega)}^2=\int_0^1 E(f^2(x,\omega)) dx$ ? – W. Volante Jun 11 '18 at 10:50
• Yes, $\int_0^1 E[\lvert f(x,\bullet)\rvert^2]\,dx$ by Fubini's theorem. – Saucy O'Path Jun 11 '18 at 10:59
• We can use Fubini because we look at the square of $f$ (ie it's positive) ? – W. Volante Jun 11 '18 at 13:20