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I know that the function

$ f(x) = \begin{cases} \frac{1}{x^2} &\mbox{if } x > 1 \\ 0 &\mbox{otherwise.} \end{cases} $

is Lebesgue integrable on $\mathbb{R}$. I want to write a function $F(x): \mathbb{R}^d \to \mathbb{R}$ such that $\int_{\mathbb{R}^d} F = \int_\mathbb{R} f$. The intuition I have is to write $F(x) = f(x)g(x)$ where $f(x)$ is defined on $\mathbb{R}$ and $g(x)$ is defined on $\mathbb{R}^{d-1}$ such that $\int_{\mathbb{R}^{d-1}}g = 1$ and use Fubini's Theorem to split up the double integral for $F$. However, I'm having trouble cooking up the exact right function for $h$.

1 Answers1

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On the unit $d$-dimensional square define $F(x)=\int_{\mathbb{R}}f$ and zero otherwise.

In this way $$\int_{\mathbb{R}^d}F=\text{(Volume of unit $d$-dimensional square)}\cdot\int_{\mathbb{R}}f$$

The part that you need to clarify is in what way $F$ must extend $f$?

If $F$ must be equal to $f$ on $(t,0,0,...,0)$ for example. You could modify the example above such that on the line $(t,0,0,...,0)$ $F=f$. This doesn't affect the value of the integral since the line has measure zero.

  • Ah, I was looking for an extension where $F(x, 0, \dots, 0) = f(x)$ as you mentioned. Even then, the construction where $F = f(x)\chi_{[0,1]^d}$ you suggested works since the value of the integrals differs only on a set of measure $0$ (the line). –  Feb 23 '15 at 20:01
  • @shuckles The construction above is not $F=f\chi_{[0,1]^d}$. It is rather $F=\chi_{[0,1]^d}\int_{\mathbb{R}}f$, it is constant on $[0,1]^d$. –  Feb 23 '15 at 20:23