Why is this integral not finite? This question is related to a previous one, I asked here, though I'm not going to explicitly state their relationship.
Let $f:\mathbb{R} \to \mathbb{R}$ be a smooth compactly supported function and let $a_1$ and $a_2$ be two non-zero real numbers with non-zero solutions to the equation $a_1x + a_2y = 0 $ for $(x,y) \in [0,\infty)^2$. I want to prove that the following integral below is not necessarily finite:
$$\int_{ [0,\infty)^2 } f(a_1x+ a_2y) \ dx \ dy $$
The thing is that I "sort of" understand the intuition of why this is so, but I'm struggling to express it rigorously (as are the texts I've read claiming this fact ...)
Let $K$ the set of $(x,y) \in [0,\infty)^2$ such that $a_1x + a_2y = 0$. Because $K$ contains a non-zero vector, we can scale that non-zero vector by positive scalars and get a new non-zero solution for each scale. Therefore, $K$ is a ray. Now let  $(x_t,y_t) \in [0,\infty)^2$ such that $a_1x_t +a_2y_t = t $ where $t$ lies in the support of $f$. Then we should have something like
\begin{align*}
\int_{ [0,\infty)^2 } f(a_1x+ a_2y) \ dx \ dy & \geq \int_{K + (x_t,y_t)} f(a_1x+ a_2y) \ dx \ dy \newline
&  = \int_{K+(x_t,y_t)} f(t) \ dx \  dy
\end{align*}
and we should have that the last integral is not finite. Of course, the last two integrals don't make sense so the inequality doesn't make sense. But it feels like it's capturing the idea. I would appreciate some help in completing my idea into a rigorous proof or by being given a totally different (and rigorous) argument.
 A: For $ax+by=0$ to have a solution for positive $x,y,$ one of $a,b$ must be positive and the other negative. Without loss of generality, assume $a<0<b.$
Then the set of solutions of $ax+by=0$ for $x,y$ non-negative are $(bt,-at)$ for $t\in[0,\infty).$
Assume $f(x)>\epsilon$ for all $x\in (c,d).$
We will first assume $c>0.$
Then let $C_{(c,d)}=\{(x,y)\in[0,\infty)^2\mid ax+by\in (c,d)\}.$ This is an open subspace of $[0,\infty)^2.$
Given any $ax_0+by_0=0$ we can get $ax_0+b(y_0+y)\in (c,d)$ for $y\in (c/b,d/b).$ So this makes $C_{(c,d)}$ an infinite rectangle past a cutoff, of slope $(b,-a)$ and diagonal width parellel to the $y$-axis equal to $\frac{d-c}{b}.$ So this means $C_{(c,d)}$ has infinite measure, and $f(ax+by)>\epsilon$ for $(x,y)\in C_{(c,d)}.$
So the integral is not finite.
If $d<0,$ then we instead choose the points $(x_0+x,y_0)$ such that $a(x_0+x)+by_0\in (c,d).$ We again get an infinite rectangle.
Finally, if $c\leq 0<d$ then we can take $(c',d')=(d/2,d)$ and use one of the cases above. If $c<0=d,$ we take $(c',d')=(c,c/2).$
The same argument works if $f(x)<-\epsilon$ for all $x\in(c,d).$
You might get some non-absolute convergence, depending on the order of integration, if $\int_{\mathbb R} f(x)\,dx =0.$ But in general, if $f$ is continuous the integral will not converge.
A: I am posting an answer inspired by Thomas Andrews's answer:
Consider the map $g:{[0,\infty)^2 \to \mathbb{R}}$ where $g: (x,y) \mapsto ax_1 + bx_2 $. If $x_t,y_t$ satisfies $g(x_t,y_t) = t \neq 0$, then we can scale $(x_t,y_t)$ by any non-negative scalar to guarantee that the image of $g$ must contain $[0,\infty)$ or $(-\infty,0]$
Choose $f\in C_{c}^{\infty}(\mathbb{R}) $ such that $f(x) > \epsilon$ for some $(c,d)$ not containing $0$ in the image of $g$. (This is possible since $f$ is continuous.)
Since $g$ is continuous and $(c,d)$ lies in the image of $g$, $g^{-1}(c,d)$ is a non-empty open subset of $[0,\infty)^2 $. Therefore, it contains an open rectangle $R$ of finite area. $g^{-1}(c,d)$ is closed under the translation of $K$ (which as mentioned, is a ray). So $K+R$ lies in $g^{-1}(c,d)$. But $K+R$ has infinite area and so
Therefore,
\begin{align}
\int_{ [0,\infty)^2 } f(a_1x+ a_2y) \ dx \ dy & \geq \int_{K + R} f(a_1x+ a_2y) \ dx \ dy \newline
&  \geq \int_{K+R} \epsilon \ dx \  dy     
\end{align}
which has infinite area.
