Set of discontinuous points Suppose $f$ is function from $\mathbb{R}$ to $\mathbb{R}$. Let be the set $\mathbf{A}$ that contains all the discontinuous points of $f$.Is  $\mathbf{A}$ Borel Measureable?
 A: The set of point of continuity is a $G_\delta$ set. So the points of discontinuity is $F_\sigma$. $F_\sigma$ sets are certainly Borel.

To see this: Let $C$ be the set of points of continuity of $f$. Define $\text{osc}_f(x) = \inf \{\text{diam}(f(U)) : x \in U \text{ and } U \text{ is open }\}$. Note $f$ is continuous at $x$ if and only if $\text{osc}_f(x) = 0$. Show that for each $\epsilon$, the set $E_\epsilon = \{x : \text{osc}_f(x) < \epsilon\}$ is open. Then $C = \{x : \text{osc}_f(x) = 0\} = \bigcap_n E_{\frac{1}{n}}$. So what you called $A = \mathbb{R} - C$ is $F_\sigma$.

The oscillation is just how small the image an open set containing $x$ can be made. Suppose $f$ is continuous at $x$, then for all $\epsilon$, there exists a $\delta$ such that $|f(x) - f(y)| < \epsilon$ whenever $|x - y| < \delta$. This means that $\operatorname{diam}((x - \delta, x + \delta)) < 2\epsilon$ (use triangle inequality). So $\operatorname{osc}_f(x) < 2\epsilon$. Since $\epsilon$ is arbitrary, $\operatorname{osc}_f(x) = 0$. Conversely suppose $\operatorname{osc}_f(x) = 0$. Then for all $\epsilon$, there exist an open set $U$ containing $x$ such that $|f(a)  - f(b)| < \epsilon$ if $a,b \in U$. Choose $\delta$ such that $(x - \delta, x + \delta) \subseteq U$. Then if $|x - y| < \delta$, $|f(x) - f(y)| < \epsilon$. So $f$ is continuous. 
