Suppose $f \in L_+^1$ let $g(x,y)= \min(f(x), f(y))$. Is $g\in L_{+}^1$?

Question

Consider a function $f \ge 0$ such that $\int f(x) dx <\infty$. Now let $$g(x,y)= \min(f(x), f(y)).$$

Question: Can we show that $$\int \int g(x,y) dx dy <\infty$$? Or is there a counter-example?

I tried playing with examples but really could find anything. My intuition tells me that to find a counter-example somehow we have to make the function $g$ blow up really fast around zero.

Answer 1

I found this counterexample:

$f(x)=\frac{1}{x^2}\chi_{[1,\infty)}\in L^1_+,$ and for $0<x<y\implies f(x)>f(y)$

so defining the set $D:=\{(x,y)|y>x>1\}$ we have $g(x,y)_{|D}=\frac{1}{y^2}$:

$$\int_{\mathbb{R}^2}|g|\geq \int_{D}|g|= \lim_{R\rightarrow \infty }\int_1^R \text dx\int_x^R \text dy \frac{1}{y^2} =\lim_{R\rightarrow \infty }\int_1^R \text dx \left[\frac{1}{x}-\frac{1}{R}\right]=\infty $$