Integrating multiple of two measurable functions

Question

Let $G(x,y): \mathbb{R}_0^+ \times \mathbb{R}_0^+ \to \{0,1,\infty\}$ be a (Borel) measurable function. If the set of values $(x,y) \in \mathbb{R}_0^+ \times \mathbb{R}_0^+ $ such that $G(x,y) = \infty$ has positive Lebesgue measure, does the following hold? $$\int_{0}^\infty\int_{0}^\infty G(x,y) \exp \{-x - y \} \, dx \, dy= \infty$$ Why or why not?

Answer 1

In the case that there is a set $A$ such that $G^{-1}(\infty)=A$ and $\mu(A)>0$ there is a $R>0$ such that $\mu(B_R(0)\cap A)>0$, where $B_R(0)$ is an open ball (in the quarter space) of radius $R$. Note that $e^{-x - y }$ is bounded from below by some $c>0$ on $B_R(0)\cap A$. Hence \begin{align*} \int\int_{\mathbb{R}_0^+ \times \mathbb{R}_0^+} G(x,y) e^{-x - y } \, dx \, dy\geq\int\int_{B_R(0)\ \cap\ A} G(x,y)\cdot c \, dx \, dy=\infty. \end{align*} Just to add: If $G$ would be bounded by $1$ outside of a set $A$ with Lebesgue measure $\mu(A)=0$ then \begin{align*} 0\leq \int\int_{\mathbb{R}_0^+ \times \mathbb{R}_0^+} G(x,y) e^{-x - y } \, dx \, dy\leq\int\int_{\mathbb{R}_0^+ \times \mathbb{R}_0^+\setminus A} 1\cdot e^{-x - y } \, dx \, dy< \infty. \end{align*}

Answer 2

Let $E$ be such a set. There is a compact set $K$ such that $m(E\cap K)>1.$ Let $\alpha=\min \{\exp \{-x - y \}:(x,y)\in K)\}$ and note that $\alpha>0.$ Now

$\int_{0}^\infty\int_{0}^\infty G(x,y) \exp \{-x - y \} \, dx \, dy=\int_{\mathbb{R}_0^+ \times \mathbb{R}_0^+} G(x,y) \exp \{-x - y \} dm\ge$

$ \int_{K\cap E} G(x,y) \exp \{-x - y \} dm= \infty$