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$\textbf{Question}$

Let $(\Omega_1, d_1)$ and $(\Omega_2, d_2)$ be metric spaces and let $f: \Omega_1 \rightarrow \Omega_2$ be an arbitrary map. Denote by $U_f = \{ x \in \Omega_1 \, | \, f ~ \textrm{is discontinuous at}~x \}$ the set of points of discontinuity of $f.$ Show that $U_f \in \mathcal{B}(\Omega_1) \triangleq$ the Borel $\sigma$-algebra on $\Omega_1.$

$\textit{Hint}$: First show that for any $\epsilon > 0$ and $\delta > 0$ the set

\begin{align*} U^{\delta, \epsilon}_{f} = \left\{ x \in \Omega_1 \, \middle| \, \textrm{there are}~ y,z \in \overset{\circ}{B}_{\Omega_1}(x; \epsilon) ~ \textrm{with}~ d_2(f(y), f(z)) > \delta \right\} \end{align*} is open, where for any $i \in \{ 1, 2 \},$ $\overset{\circ}{B}_{\Omega_i}(z; \rho) = \{ y \in \Omega_i \, | \, d_{i}(z,y) < \rho \}$ with $(z, \rho) \in \Omega_i \times \,]0,+\infty[.$

$\textbf{Solution Attempt}$:

For any $i \in \{ 1, 2 \},$ let $B_{\Omega_i}(z; \rho) = \{ y \in \Omega_i \, | \, d_{i}(z,y) \leq \rho \}$ with $(z, \rho) \in \Omega_i \times \,]0,+\infty[$ (i.e., the closed ball). Then $U^{\delta, \epsilon}_{f}$ can be written as \begin{align*} U^{\delta, \epsilon}_{f} = \left\{ x \in \Omega_1 \, \middle| \, (\exists y \in \overset{\circ}{B}_{\Omega_1}(x; \epsilon))(\exists z \in \overset{\circ}{B}_{\Omega_1}(x; \epsilon)) ~~~~ f(y) \in \complement_{\Omega_2} B_{\Omega_2}(f(z); \delta) \right\}, \end{align*} where $\complement_{\Omega_2}$ denotes the complement of a set with respect to $\Omega_2.$ Hence, it suffices to show that $\complement_{\Omega_1} U^{\delta, \epsilon}_{f}$ is closed in $\Omega_1.$ To that end, observe that \begin{align*} \complement_{\Omega_1}U^{\delta, \epsilon}_{f} = \left\{ x \in \Omega_1 \, \middle| \, (\forall y \in \overset{\circ}{B}_{\Omega_1}(x; \epsilon))(\forall z \in \overset{\circ}{B}_{\Omega_1}(x; \epsilon)) ~~~~ f(y) \in B_{\Omega_2}(f(z); \delta) \right\}, \end{align*} This set is closed, as for any sequence $(x_n)_{n \in \mathbb{N}}$ in $\complement_{\Omega_1}U^{\delta, \epsilon}_{f},$ we have \begin{align*} (\forall n \in \mathbb{N})(\forall y \in \overset{\circ}{B}_{\Omega_1}(x_{n}; \epsilon))(\forall z \in \overset{\circ}{B}_{\Omega_1}(x_{n}; \epsilon)) ~~~~ f(y) \in B_{\Omega_2}(f(z); \delta); \end{align*} i.e., $\underset{n \rightarrow +\infty}{\lim} \, x_n \in \complement_{\Omega_1}U^{\delta, \epsilon}_{f}.$ Thus, $U^{\delta, \epsilon}_{f}$ is open in $\Omega_1.$ Set $U^{\delta}_{f} = \bigcup\limits_{\epsilon \in \,]0,+\infty[} U^{\delta, \epsilon}_{f}.$ Then $U^{\delta}_{f}$ is also open, as it is an arbitrary union of open sets. Finally, we observe that $U_{f} = \bigcup\limits_{\delta \in \,]0,+\infty[} U^{\delta}_{f} = \bigcup\limits_{m \in \mathbb{N}} U^{\frac{1}{m+1}}_{f}.$ As a countable union of open sets, we can conclude that $U_f \in \mathcal{B}(\Omega_1).$ $\square$

Trouble Is Back
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1 Answers1

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For $f:\Omega_1\to\Omega_2$ , oscillation of $f$ is defined at a point $x \in \Omega$ by $\omega_f(x) =\inf \{\operatorname{diam}(f(U)): x\in U \in \tau_{d_1}\}$

Then $f$ is discontinuous at $x$ iff $\omega_f(x)>0$

The set of all discontinuity of $f$ , $D_f=\{x\in \Omega : \omega_f(x)>0\}$

Given $\epsilon>0$ , $D_f(\epsilon) =\{x\in \Omega : \omega_f(x)\ge \epsilon\}$

$\omega_f(\epsilon)$ is closed set for each $\epsilon>0$ (see here or $\omega_f $ is upper semi-continuous here)

Then show that $D_f=\bigcup_n D_f(\frac{1}{n}) $

Then $D_f$ is a $F_{\sigma}$ set, hence is in $\mathcal{B}(\Omega_1) $

Sourav Ghosh
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  • This looks similar to what I initially wrote. Except that I think that I need to define $U^{\delta}{f} = \bigcap\limits{\epsilon \in ,]0,+\infty[} U^{\delta, \epsilon}{f} = \bigcap\limits{n \in \mathbb{N}} U^{\delta, \frac{1}{n+1}}_{f}.$ Is this correct? – Trouble Is Back Jul 09 '22 at 04:57