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Having $f^2$ integrable over $R$, with $f:R\subset \mathbb{R}^{n}\rightarrow \mathbb{R}$ bounded. How does one apply the known result from unidimensional calculus? (If $f^2$ integrable, then $|f|$ is integrable?). Over the given domain in a rectangle.

PeterP123
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  • I think you want to say that $R$ is bounded, not $f$. – Robert Israel Oct 29 '21 at 17:30
  • Much more generally, if $g:D\to \mathbb{R}$ is continuous, $h:R \to \mathbb{R}$ is Riemann integrable,, and $h(R) \subset D$, then $g\circ h$ is Riemann integrable. Here $g(x) = x^{1/2}$ and $h(x) = |f(x)|^2$. The proof here for the 1-D case generalizes easily to the multidimensional case. – RRL Oct 29 '21 at 17:38
  • Are you asking about Riemann or Lebesgue integrals here? – RRL Oct 29 '21 at 19:21

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One of the answers in the linked question uses the analogue of the inequality $|f(x,y)| \le 1 + f(x,y)^2$ which is valid for any real number $f(x,y)$. Basic properties of the integral give you $$\int_R |f(x,y)| \, dxdy \le \int_R 1 \, dxdy + \int_R f(x,y)^2 \, dxdy$$ and your hypothesis is that the right hand side is finite.

Umberto P.
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  • But isn't this just proving it for the particular case of $\mathbb{R}^2$, the concern is if this works undistinguished for all $\mathbb{R}^n$. – PeterP123 Oct 29 '21 at 18:03
  • I think once you convince yourself that the two-dimensional case doesn't depend on the corresponding one-dimensional result you'll see that it is valid in any number of dimensions. – Umberto P. Oct 29 '21 at 18:57