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 $\infty$).

The other thing does not make sense.

  • $\begingroup$ So $\lVert f\rVert_{L^2(]0,1[\times\Omega)}^2=\int_0^1 E(f^2(x,\omega)) dx$ ? $\endgroup$ – W. Volante Jun 11 '18 at 10:50
  • $\begingroup$ Yes, $\int_0^1 E[\lvert f(x,\bullet)\rvert^2]\,dx$ by Fubini's theorem. $\endgroup$ – Saucy O'Path Jun 11 '18 at 10:59
  • $\begingroup$ We can use Fubini because we look at the square of $f$ (ie it's positive) ? $\endgroup$ – W. Volante Jun 11 '18 at 13:20
  • $\begingroup$ Yes. Technically that version would be Tonelli's, but whatever. $\endgroup$ – Saucy O'Path Jun 11 '18 at 13:27

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