If we have two functions $f,g$ defined on a region $R$ such that $f(x,y) \leq g(x,y),$ for all $(x,y) \in R,$ we can guarantee that $$ \iint_R f(x,y) dA \leq \iint_R g(x,y)dA. $$ This is a well-known fact.
Now, assume that we can also guarantee that $$ \iint_R f(x,y)dA = +\infty$$ Can we also guarantee that $$ \iint_R g(x,y)dA \overbrace{=}^{?} +\infty $$ If so, what is the reasoning behind this? Sometimes I get quite confused when it comes to infinity.
Thanks for any help in advance.
Formally, $$ \iint_R f(x,y)dA = +\infty$$ means something like "For an arbitrarily large number $N\in\mathbb{R}_{>0}$ there is some partition of $R$ into smaller parts $R_1,\dots,R_n$ (such that $R=\cup_i R_i$) and numbers $\lambda_1,\dots,\lambda_n$ such that $f(x)>\lambda_i$ for $x\in R_i$, and $N<\sum_i \lambda_i A(R_i)$, where $A(R_i)$ is the area of the region $R_i$."
(The precise definition depends on your type of integration. esp. using Riemann integration, you should also split out positive and negative, and then the regions $R_i$ should be rectangles.)
The point here is that $\sum_i \lambda_i\mathbf{1}_{R_i}$ is an approximation of $f$ by a simple function, and in particular a lower bound.
Although this is a bit technical, your question quickly follows: When $f\leq g$, then for any big $N$ and any "approximation" $\sum_i\lambda_i\mathbf{1}_{R_i}$ of $f$ such that $N<\sum_i\lambda_iA(R_i)$, we also have that $ g(x)>\lambda_i$ for $x\in R_i$, so this "approximation" also approximates/is a lower bound of $g$, so indeed $$ \iint_R g(x,y)dA = +\infty$$