Proof: Define a real-valued function $g$ on $[a, b]$ by $g(x) = \max\{|f(x) - f(x+)|, |f(x) - f(x-)|\}$. Then $f$ is continuous at $x$ iff $g(x) = 0$. Let $D_n = \{x \in [a, b]: g(x) \geq 1/n\}$. The points of discontinuity of $f$ is exactly $\bigcup_{n = 1}^\infty D_n$. We prove that each $D_n$ is finite.
Suppose on the contrary that $D_n$ is infinite. Then there is a sequence $\{x_n\}$ of distinct points in $D_n$ which converges to $x \in [a, b]$. Since $f(x+), f(x-)$ exist, we can choose some $\delta > 0$ such that $|f(y) - f(x+)| < 1/4n$ whenever $0 < y - x < \delta$ and $|f(y) - f(x-)| < 1/4n$ whenever $0 < x - y < \delta$. Therefore $|f(y) - f(z)| < 1/2n$ whenever $y, z \in (x - \delta, x)$ or $y, z \in (x, x + \delta)$. We can find some $k$ such that $0 < |x_k - x| < \delta$. If $x_k > x$, then $|f(y) - f(x_k)| < 1/2n$ whenever $y \in (x, x + \delta)$, let $y \rightarrow x_k$ from both left and right we get $|f(x_k+) - f(x_k)| \leq 1/2n$ and $|f(x_k-) - f(x_k)| \leq 1/2n$, so $g(x) \leq 1/2n < 1/n \leq g(x)$, contradiction. If $x_k < x$ then similarly we also obtain a contradiction. Hence $D_n$ must be finite. QED.
Proposition. Let $f_n: U \rightarrow \mathbb{R}$ be a sequence of continuous functions on $U \subseteq \mathbb{R}^d$ such that $f_1 \leq f_2 \leq \cdots$ and $\{f_n\}$ converges pointwise to a continuous function $f$ on $U$. Then $f_n \rightarrow f$ uniformly on compact subsets.
Proof: Note that $f_1 \leq \cdots \leq f_n \leq \cdots \leq f$. Let $K \subseteq U$ be compact and $\epsilon > 0$. For any $x \in K$, since $f_n(x) \rightarrow f(x)$, there is $N_x \in \mathbb{N}$ such that $f(x) - f_n(x) < \epsilon/3$ whenever $n \geq N_x$. For any $n \in \mathbb{N}$, by uniform continuity of $f$ and $f_n$ on $K$, there is $\delta_n > 0$ such that for any $x, y \in K$, $|f(x) - f(y)| < \epsilon/3$ and $|f_n(x) - f_n(y)| < \epsilon/3$ whenever $|x - y| < \delta_n$. For every $x \in K$, if $y \in B(x, \delta_{N_x})$, then
$\begin{align*}
&f(y) - f_n(y) \leq f(y) - f_{N_x}(y) \\ &= [f(y) - f(x)] + [f(x) - f_{N_x}(x)] + [f_{N_x}(x) - f_{N_x}(y)] \\ &< \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon
\end{align*}$
whenever $n \geq N_x$. Since $\{B(x, \delta_{N_x}): x \in K\}$ is an open cover of $K$, then there are $x_1, \ldots, x_k \in K$ such that $K = \bigcup_{j = 1}^k B(x_j, \delta_{N_{x_j}})$. Take $N = \max\{N_{x_1}, \ldots, N_{x_k}\}$. For $n \geq N$, if $x \in K$, then $x \in B(x_j, \delta_{N_{x_j}})$ for some $1 \leq j \leq k$, so $f(x) - f_n(x) < \epsilon$. Hence $f_n \rightarrow f$ uniformly on $K$. QED.
(To be continued)
Proposition. Let $f_n: U \rightarrow \mathbb{R}$ be a sequence of continuous functions on $U \subseteq \mathbb{R}^d$ such that $f_1 \leq f_2 \leq \cdots$ and $\{f_n\}$ converges pointwise to a continuous function $f$ on $U$. Then $f_n \rightarrow f$ uniformly on compact subsets.
Proof: Note that $f_1 \leq \cdots \leq f_n \leq \cdots \leq f$. Let $K \subseteq U$ be compact and $\epsilon > 0$. For any $x \in K$, since $f_n(x) \rightarrow f(x)$, there is $N_x \in \mathbb{N}$ such that $f(x) - f_n(x) < \epsilon/3$ whenever $n \geq N_x$. For any $n \in \mathbb{N}$, by uniform continuity of $f$ and $f_n$ on $K$, there is $\delta_n > 0$ such that for any $x, y \in K$, $|f(x) - f(y)| < \epsilon/3$ and $|f_n(x) - f_n(y)| < \epsilon/3$ whenever $|x - y| < \delta_n$. For every $x \in K$, if $y \in B(x, \delta_{N_x})$, then
$\begin{align*}
&f(y) - f_n(y) \leq f(y) - f_{N_x}(y) \\ &= [f(y) - f(x)] + [f(x) - f_{N_x}(x)] + [f_{N_x}(x) - f_{N_x}(y)] \\ &< \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon
\end{align*}$
whenever $n \geq N_x$. Since $\{B(x, \delta_{N_x}): x \in K\}$ is an open cover of $K$, then there are $x_1, \ldots, x_k \in K$ such that $K = \bigcup_{j = 1}^k B(x_j, \delta_{N_{x_j}})$. Take $N = \max\{N_{x_1}, \ldots, N_{x_k}\}$. For $n \geq N$, if $x \in K$, then $x \in B(x_j, \delta_{N_{x_j}})$ for some $1 \leq j \leq k$, so $f(x) - f_n(x) < \epsilon$. Hence $f_n \rightarrow f$ uniformly on $K$. QED.
(To be continued)
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