The integral on $[0,1]\times[0,1]$ Here I have a problem. 
$p$ and $q$ are positive numbers. the integral $$\int_0^1\int_0^1 \frac{1}{x^p+y^q}\;dx\;dy< \infty \Longleftrightarrow \frac{1}{p}+\frac{1}{q}>1$$
Here is my try, maybe we can change variable. When $p>1:$ $$\int_0^1\int_0^1 \frac{1}{x^p+y^q}dxdy=\int_0^1\int_0^1 \frac{1}{(\frac{x}{y^{\frac{q}{p}}})^p+1}\cdot \frac{1}{y^q}dxdy=\int_0^1\int_0^{y^{-\frac{q}{p}}} \frac{y^{\frac{q}{p}-q}}{t^p+1}dtdy$$
So when $p>1,$ integrable$\Longleftrightarrow \frac{q}{p}-q>-1\Longleftrightarrow \frac{1}{p}+\frac{1}{q}>1 $ When $0<p \leq 1, \ \text{we have}\frac{1}{x^p}\geq \frac{1}{x^p+y^q}\geq \frac{1}{(x+y^\frac{q}{p})^p}$
So integrable$\Longleftrightarrow \frac{1}{p}+\frac{1}{q}>1$
 A: Here is a more estimate-oriented and Lebesgue-oriented approach, based on two facts:


*

*The sum of two nonnegative numbers is comparable to their maximum. Namely, $\max(a,b)\le a+b\le 2\max(a,b)$.  

*The integral of a nonnegative function $f$ with respect to measure $\mu$ is equal to $\int_0^\infty \mu(\{f>t\})\,dt$.


If the above are taken for granted, then it remains to observe that for $t>1$, the set where 
$ \frac{1}{\max(x^p,y^q)} >t $
is the rectangle of dimensions $t^{-1/p}$ and $t^{-1/q}$. Its area is $t^{-(1/p+1/q)}$, the integral of which converges on $[1,\infty)$ ... you know when. 
We don't actually need the precise form of Fact 2, the following "rough" form is enough. Since $\lfloor f \rfloor \le f\le \lfloor f \rfloor+1$, the integral of nonnegative $f$ on a finite measure space converges if and only if the integral of $\lfloor f \rfloor$ converges. The latter integral is simply the sum 
$$ \sum_{k=1}^\infty \mu(\{f\ge k\})     
$$
Plugging in $f(x,y)=\frac{1}{\max(x^p,y^q)}$ we get 
$$ \sum_{k=1}^\infty \frac{1}{k^{1/p+1/q}}     
$$
