Showing the set is closed and the set of discontinuities of $f$ is possibly a union of countably many closed sets

Question

Let $f:[a,b] \to \mathbb{R}$ be a bounded function. Define $A_{f,x} : (0, \infty) \to \mathbb{R}$ by $$A_{f,x} (r) = \text{diam}\left( f((x-r,x+r) \cap [a,b])\right).$$ Show that for all $\epsilon >0$ the set $\{x\in [a,b] : \lim_{r\to 0^+} A_{f,x} (r)\geq \epsilon \}$ is a closed set.

I showed that $f$ is continuous at $x$ if and only if $\lim_{r\to 0^+} A_{f,x} (r) = 0$. I'm thinking about showing the set is closed by showing the image of the set is closed (since $f$ is continuous, the pre-image must be closed). Is this the right approach? If so/not, how should I proceed/conclude? (In this case, would the set of discontinuities of $f$ be a union of countably many closed sets?)

Answer 1

Let $$ S_\varepsilon=\{x\in [a,b] : \lim_{r\to 0^+} A_{f,x} (r)\geq \varepsilon \} $$ and let $x_0\not\in S_\varepsilon$. This means that $$ \lim_{r\to 0^+} \text{diam}\big(\, f\big[(x_0-r,x_0+r) \cap [a,b]\big]\big)< \varepsilon $$ and hence, there exists an $r_0>0$, such that $$ \text{diam}\big(\, f\big[(x_0-r_0,x_0+r_0) \cap [a,b]\big]\big)< \varepsilon $$ Claim. If $x\in (x_0-r_0,x_0+r_0)$, then $\lim_{r\to 0^+}A_{f,x}<\varepsilon$.

Indeed, if $x_1\in (x_0-r_0,x_0+r_0)$, then for $r_1=|r_0-x_1|$, we have that $$ (x_1-r_1,x_1+r_1)\subset (x_0-r_0,x_0+r_0) $$ and hence $$ \text{diam}\big(\, f\big[(x_1-r_1,x_1+r_1) \cap [a,b]\big]\big)< \varepsilon $$ which in turn implies that $$ \lim_{r\to 0^+} \text{diam}\big(\, f\big[(x_1-r,x_1+r) \cap [a,b]\big]\big)< \varepsilon. $$

Therefore, if $x_0\not\in S_\varepsilon$, then a whole open interval containing $x_0$ is disjoint with $S_\varepsilon$. This means that the complement of $S_\varepsilon$ is open and hence $S_\varepsilon$ is closed.

Answer 2

It is easier to prove that the complement is open. Suppose $lim J_{f,x} (r) < \epsilon$. Let $lim J_{f,x} (r) <\epsilon ' <\epsilon$. Then there exists $\delta >0$ such that $J_{f,x} (r) < \epsilon '$ for $0<r<\delta$. Let $|x-y|< \frac {\delta} 4$. Then $(y-t,y+t)$ is a subset of $(x-\delta /2,x+ \delta /2)$ provided $0<t<\delta /4$. Hence $J_{f,y} (t) \leq \epsilon ' $ for $0<t<\delta /4$. This proves that $lim J_{f,y} <\epsilon$. This holds for all $y$ such that $|x-y|< \frac {\delta} 4$, proving that the complement is open.

Answer 3

We don't know that $f$ is continuous. What we only have assumed is $f$ is bounded. Your set which will be shown to be closed is a set of discontinuities, and there is no reason that the set is empty.

Here is another way to prove your set is closed: we will use an important characterization of closed set, that is, a set is closed iff it is closed under the limit of sequences over a set. For convenience let $g(x) = \lim_{r\to 0^+} A_{f,x}(r)$. $g(x)$ is well defined as $A_{f,x}(r)$ is a bounded increasing function over $(0,\delta)$ for some small $\delta>0$.

Let $\langle x_n : n\in\mathbb{N}\rangle$ be a sequence such that $g(x_n)\ge\varepsilon$ for all $n$ and has a limit $x$. We shall check that $g(x) \ge\varepsilon$. For each $n$ choose $\langle r_{n,m}: m\in\mathbb{N}\rangle$ such that

1. $r_{n,m}\le 1/n$, $\lim_m r_{n,m}=0$ and
2. $A_{f,x_n}(r_{n,m}) \ge g(x_n) - 1/n$ as $m\to\infty$.

Moreover, we can also choose $u_{n,m}, v_{n,m}\in (x-r_{n,m},x+r_{n,m})$ such that $$|f(u_{n,m})-f(v_{n,m})| \ge A_{f,x_n}(r_{n,m}) - \frac{1}{n}.$$

We can see that $|x-u_{n,m}|$ and $|x-v_{n,m}|$ are less than $|x-x_n| + r_{n,m}$, hence $$A_{f,x}(r_{n,m}+|x-x_n|) \ge A_{f,x_n}(r_{n,m})-\frac{1}{n} \ge \varepsilon - \frac{2}{n}.$$

Take a limit for $n\to\infty$ then the left-hand-side converges to $g(x)$ and the right-hand-side is $\varepsilon$.