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?
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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.
Martin Sleziak
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William
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can u explain in words what that osc means? and why is osc=0 iff f continious. – user95525 Oct 08 '13 at 07:27
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@user95525 I showed above that $C$ is $G_\delta$, which means it is a countable intersection of open sets. $G_\delta$ sets are Borel. You do know that the collection of Borel sets contain the open sets and are closed under complements and countable intersection? – William Oct 08 '13 at 08:22
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Im following the course Measure theory and i find it very difficult. I forgot that the collection of Borel sets contain the open sets and are closed under complements and countable intersection. – user95525 Oct 08 '13 at 08:24
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@user95525 You can find some basic things about oscillation of a function on Wikipedia. Maybe that can help you, too. – Martin Sleziak Oct 08 '13 at 09:08
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@William can you please explain why is $E_\epsilon = {x : \text{osc}_f(x) < \epsilon}$ open ? I know that inverse image of an open set is open for continuous function, but here I think we don't know whether $\text{osc}_f(x)$ is continuous or not. – Mathronaut Sep 18 '14 at 19:12
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@NeerajBhauryal In metric spaces, a set is open if every point in the set has an open ball contained in the set. Use this definition to prove the set is open. – William Sep 18 '14 at 23:33
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@William could you please show how it is done, I'm not able to do it. Moreover if we show it is true for all $\epsilon$ then won't this imply that $\text{osc}_f$ is continuous ? which is not true in general! – Mathronaut Sep 19 '14 at 07:24
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@William thanks for the answer. A small typo. It was writtten as $\operatorname{diam}((x - \delta, x + \delta)) < 2\epsilon$ but should be $\operatorname{diam}(f(x - \delta, x + \delta)) < 2\epsilon$. – nrynn Oct 23 '17 at 12:57
youinstead ofu; if yes, then this is 2013 and surely you can find a better deal.) – Asaf Karagila Oct 08 '13 at 07:32