Doubt over a real analysis problem Consider a measurable function $h$ from real number to real number. fix $\delta,\epsilon>0$. Consider the set $E$ of $x \in \Bbb R$ for which $\exists y$ s.t. $|x-y|<\delta$ and $|h(y)-h(x)|\geq \epsilon$. Is this true that $E$ is open ? 
 A: If I am not mistaken the set will not be open in general. Consider the following function $h : \mathbb{R} \to \mathbb{R}$.
$$ h(x) = \begin{cases} 1 & \text{ if } x \geq 0\\ 0 &\text{ if } x \leq 0 \land x \in \mathbb{Q} \\ \frac{1}{2} & \text{ otherwise } \end{cases} $$
Let $\epsilon = \frac{2}{3}$ and $\delta = 1$. First we have $-\frac{1}{2} \in E$ since $h\left(-\frac{1}{2}\right) = 0$ and so we may pick $y$ to be $0$ to have $\left|h\left(-\frac{1}{2}\right) - h(y)\right| = 1 \geq \epsilon$.
Second, no negative irrational number is in $E$. Indeed, let $x < 0$ be irrational. Then $h(x) = \frac{1}{2}$. Given any $y$, we will either have $|h(x) - h(y)| = \frac{1}{2}$, or $|h(x) - h(y)| = 0$ and none of these are greater or equal to $\epsilon = \frac{2}{3}$. 
Hence no neighbourhood of $-\frac{1}{2}$ is in $E$, so $E$ is not open.
Lastly, the function $h$ is measurable since it is a sum of two characteristic functions $h = \chi_{A_1} + \frac{1}{2}\chi_{A_2}$ where the sets $A_1$ and $A_2$ are $$A_1 = \{ x \in \mathbb{R}\ |\ x \geq 0 \}$$ $$A_2 = \{ x \in \mathbb{R}\ |\ x \leq 0 \land x \in \mathbb{R}\setminus\mathbb{Q}\}$$ both of which are basic examples of measurable sets.
Of course sometimes the set will be open. For instance if the function is constant the set is empty.
