Basic Measure Theory: Borel ##\sigma##- algebra

[WMDhamnekar]

TL;DR Summary

Let $(\Omega_1, d_1)$ and $(\Omega_2, d_2)$ be metric spaces and let $f : \Omega_1 \to \Omega_2$ be an arbitrary map. Denote by $U_f =\{x \in \Omega_1 :$ f is discontinuous at $x\}$ the set of points of discontinuity of $f$. Show that $U_f \in \mathcal{B}(\Omega_1).$

My answer:
We can proceed as follows. Let $x \in U_f$. Since $f$ is discontinuous at $x$, there exists $\varepsilon > 0$ such that for some $\delta > 0$, we can find $y, z \in B_{\varepsilon}(x)$ with $d_2(f(y),f(z)) > \delta$. Therefore, $x \in U^{\delta,\varepsilon}_f$. So, $U_f \subset U^{\delta,\varepsilon}_f$. Since $U^{\delta,\varepsilon}_f$ is open, any subset of it is also open. So $U_f$ is open.

Next, we show that for any $\delta, \varepsilon>0$, we have $U^{\delta, \varepsilon}_f \subset U_f$. Let $x \in U^{\delta, \varepsilon}_f$. Then there exist $y,z \in B_{\varepsilon}(x)$ such that $d_2(f(y),f(z)) > \delta$. Since $f(y) \neq f(z)$, $f$ cannot be continuous at $x$. Therefore, $x \in U_f$. Thus, $U^{\delta,\varepsilon}_f \subset U_f$.

Taking intersection over all $\delta$, $\varepsilon$, we have $U_f = \bigcap_{ \delta, \varepsilon > 0}U^{\delta,\varepsilon}_f$. Since intersections of open sets are open, $U_f$ is open.
Here is an example:

Let $\Omega_1 = \mathbb{R}$ with the usual metric $d_1(x,y) = |x-y|$, and $\Omega_2 = \mathbb{R}^2$ with the Euclidean metric $d_2((x_1,y_1),(x_2,y_2)) = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$.

Define $f:\mathbb{R} \to \mathbb{R}^2$ as $f(x) = (x, x^2)$ if $x \neq 0$ and $f(0) = (0,1)$.

Then, $f$ is discontinuous at $x=0$. For any $\varepsilon > 0$, taking $\delta = 1$, we can find $y = -\varepsilon/2 < 0$ and $z = \varepsilon/2 > 0$ in $B_{\varepsilon}(0)$ such that $d_2(f(y),f(z)) = |(-1/2,0) - (1/2,0)| = 1 > \delta = 1$. Therefore, $0 \in U^{\delta,\varepsilon}_f$ for all $\delta, \varepsilon > 0$. Hence, $0 \in U_f$.

Also, for any $x \neq 0, f$ is continuous at $x$. Therefore, $U_f = \{0\}$. Since $\{0\}$ is open in $\mathbb{R}$, $U_f$ is open.
Hence, in this example, $U_f = \{0\} \in \mathcal{B}(\Omega_1 = \mathbb{R})$.

Therefore, $U_f$ is an open subset of $\Omega_1$. Hence, $U_f \in \mathcal{B}(\Omega_1)$.

Is this answer and example correct?

 

[member38644]

Yes, your answer and example are correct. You have correctly shown that $U_f$ is an open subset of $\Omega_1$ and therefore belongs to the Borel $\sigma$-algebra on $\Omega_1$. Your example also illustrates the concept of continuity and discontinuity in a clear and concise manner. Well done!