A problem in n-dimesnsional Lebesgue measure

Question

I found myself completely at loss while dealing with the following problem in measure theory (specificly in the field of product-measures):

Given a finite sequence $\{a_1,...,a_n\}\in\mathbb{R}$ of positive numbers, show that the function $f(\vec x)=\frac{1}{x_1^{a_1}+...+x_n^{a_n}}$ over $\mathbb{R}^n$ satisfies $$\int_{{[0,\infty)}^n\setminus[0,1]^n}f d\lambda<\infty$$

if and only if $\sum_{k=1}^n \frac{1}{a_k}<1$.

Any help would be appreciated.

Thanks in advance.

Answer 1

For $N=1,2,\dots$ let $S_N$ be the set of $x$ with $x_j>0$ for all $j$ and $$2^N\le x_1^{a_1}+\dots+x_n^{a_n}<2^{N+1},$$so that $$f(x)\sim 2^{-N}\quad(x\in S_N).$$It follows that that the integral in question is finite if and only if $$\sum_{N=1}^\infty2^{-N}m(S_N)<\infty$$

Note that $$S_{N+1}=T(S_N),$$ where $$Tx=(2^{1/a_1}x_1,\dots,2^{1/a_n}x_n).$$ Hence $$m(S_N)=c2^{N\sum\frac1{a_j}}$$(by induction), and so $\sum 2^{-N}m(S_N)<\infty$ if and only if $\sum\frac1{a_j}<1$.

Answer 2

Let $A=[0,\infty)^n\backslash[0,1]^n$.

$\Leftarrow]$ Suppose $\sum{1/a_k}<1$. Then, $a_k>1$ for all $k$. Then, by Tonelli's Theorem, the integral is finite (here, we used that $\int_{(1,\infty)}1/x^pdx<\infty$ for $p>1$).

$\Rightarrow]$ Suppose $\sum{1/a_k}\ge 1$. Then, $1/a_k\ge 1$ for some $k$. Again, by Tonelli's Theorem, we can suppose $k=1$. Then, $\int_1^\infty f(x_1,...,x_n)d\lambda(x_1)=\infty$, and thus, all the integral is infinity.