Im currently working on a proof regarding APARCH-models, where the unction $h$ in the following context comes from:
Given is the set $J^n := \{ (x,y)\in \mathbb{R}^2: |x| > m_n, y = h(x)\}$, where $m_n$ is a positive, monotonely decreasing sequence with $m_n \rightarrow 0$ and $h(x) = (|x| - \gamma x)^\delta$ for some $|\gamma| < 1$ and $\delta > 0$.
To use a nice theorem, I need (Borel-)measurability of the set $J^n$. I thought about writing the set as $J^n = (-\infty,m_n) \cup (m_n,\infty) \times h((-\infty,m_n) \cup (m_n,\infty))$, where $(-\infty,m_n)$ and $(m_n,\infty)$ as open interval are clearly measurable, hence their union is measurable too. Since $h$ is continuous, I thought that $h((m_n,\infty))$ is also open and hence measurable, but continuity only means that the preimage of an open set is open, not the image of an open set. Moreover, I don't know whether the image then is an open interval.
Moreover, I need that $\bigcup_{n=1}^\infty J^n = \mathbb{R}^2 \setminus \{0\}$, which is (clearly) not the case since $h$ is nonnegative and hence the second component of the elements in $J^n$ will never be less than $0$, so I will never reach the whole $\mathbb{R}^2$. But just out of curiosity I would like to know how I can show that the set $J^n$ is measurable (if it is at all).
Since $(x,y)\to |x|$ is continuous, the set $$A_n=\{(x,y)\in\mathbb{R}^2: |x|>m_n\}$$ is measurable for each natural number $n$.
The set $$B=\{(x,y)\in\mathbb{R}^2: y=h(x)\}$$ is a graph of a continuous function, hence it is measurable. In fact, the graph of any measurable function is measurable.
It follows that for every $n$, the set $$J^n=A_n\cap B=\{(x,y)\in\mathbb{R}^2: |x|>m_n, y=h(x)\}$$ is measurable, because it is an intersection of two measurable sets.