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 $h$.
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.