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Let $f$ be a real valued function on $\mathbb{R}$. Consider the functions
$$ w_j(x) := \sup\left\{\vert f(u) − f (v)\vert : u,v \in \left[x−\frac{1}{j},x+\frac{1}{j}\right]\right\} $$ where $j$ is a positive integer and $x\in\mathbb{R}$. Then define $$ A_{j,n}: = \left\{x \in \mathbb{R}: w_j(x)<\frac{1}{n}\right\},\qquad n=1,2,\ldots $$ and $$ A_n:=\bigcup_{j=1}^\infty A_{j,n}, \qquad n = 1,2,\ldots $$
Now let $C=\{x \in \mathbb{R}:f \text{ is continuous at }x \}$. Express $C$ in terms of the sets $A_n$.

I am totally confused. Please help me.

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What have you tried? Perhaps you can start by noting that if $x$ is a point of continuity, then $\lim_{j\rightarrow\infty}w_j(x) = 0$. Can you prove that this implies: for every $n$ there exists $j$ such that $x\in A_{j,n}$? If the last assertion holds, then $C$ must be a subset of $\cap_n A_n$. Then try to show the other inclusion. – William Jan 1 '13 at 12:28
will you explain please.i am still not getting it. – abdakchi Jan 1 '13 at 13:37

Let $x \in C$, i.e. $f$ is continuous in $x$. We want to show that $$x \in \bigcap_{n \in \mathbb{N}} A_n \tag{1}$$

Let $\varepsilon>0$. By definition of continuity there exists $\delta(\varepsilon)>0$ such that for all $y \in \mathbb{R}$, $|y-x| \leq \delta(\varepsilon)$ $$|f(y)-f(x)| \leq \varepsilon$$

Now let $u,v \in \mathbb{R}$ such that $|u-x|< \delta(\varepsilon)$, $|v-x| \leq \delta(\varepsilon)$. By triangular inequality:

$$|f(u)-f(v)| \leq |f(u)-f(x)|+|f(x)-f(v)| \leq 2 \varepsilon$$

In particular we can choose $\varepsilon := \frac{1}{2n}$, then $$|f(u)-f(v)| \leq \frac{1}{n}$$ for all $u,v$ such that $|u-x|< \delta(\frac{1}{2n})$, $|v-x| \leq \delta(\frac{1}{2n})$. Now choose $j(n) \in \mathbb{N}$ large enough such that $\frac{1}{j(n)} \leq \delta \left( \frac{1}{2n} \right)$. Then you can easily see that $x \in A_{j(n),n}$. This works for all $n \in \mathbb{N}$, so $(1)$ is proved.

Now let $x \in \bigcap_{n \in \mathbb{N}} A_n$. We want to prove that $f$ is continuous in $x$ (thus $x \in C$). Use the definitions and similar argumentation as above (So let $\varepsilon>0$. Then there exists $n \in \mathbb{N}$ such that $\frac{1}{n} \leq \varepsilon$ and since $x \in A_n$ ... )

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