1
$\begingroup$

Consider the following old exam question on measure theory:

Let $a \in \mathbb{R}$ and define $f:[0,1] \times [0,1] \to \mathbb{R}$ by $$f(x,y) := \begin{cases} (x - 1/2)^{-3} \quad 0 < y < |x - 1/2|^a\\ 0 \qquad \qquad \>\>\>\text{else}\end{cases}$$ For which values of $a$ is $f$ Lebesgue integrable?

Those exercises are mostly based on an application of Fubini. For this we first consider $|f|$, and then if we apply Fubini, we find for example $a > 2$. My question is, in this particular example how do I see that $f$ is measurable?

  • 0
    If $U\in \mathbb R$ is open then $f^{-1}(U)=f^{-1}(U\cap (\mathbb R\setminus \left \{ 0 \right \}))\cup f^{-1}(U\cap \left \{ 0 \right \}),\ $ which is either the union of an open set and an open set (if $\ U\cap \left \{ 0 \right \}=\emptyset),\ $ or the union of an open set and a closed set (if $0\in U.)$ – 2017-02-09
  • 1
    It is a measurable function multiplied by the indicatior function of a Borel set (in fact an open set). Simple as that. – 2017-02-09

0 Answers 0