if $f:[0,1] \to \mathbb{R}$ is increasing, show that $f$ is the pointwise limit of a sequence of continuous functions over $[0,1]$

Let $\mathcal{D}$ be the set of discontinuities of $f$. We know that $\mathcal{D}$ is at most countable, so we enumerate this set by $\mathcal{D} = \{ x_0, x_1, \cdots \}$. Now for each $n \geq 0$, let

$$ \Pi_n = \{ \tfrac{k}{2^n} : 0 \leq k \leq 2^n \} \cup \{ x_0, \cdots, x_n \} $$

and define $f_n : [0, 1] \to \mathbb{R}$ as the linear interpolation of the points $\{ (x, f(x)) : x \in \Pi_n\}$ ordered from left to right. Then

• It is clear that $f_n$ is continuous and increasing for each $n\geq 0$.

• If $x \in \cup_{n\geq 0} \Pi_n$, then $x \in \Pi_N$ for some $N$ and hence by construction, $f_n(x) = f(x)$ for all $n \geq N$. So we have $f_n(x) \to f(x)$ as $n\to\infty$.

• If $x \in [0, 1] \setminus \mathcal{D}$, then for each fixed $m$ there is $a_m, b_m \in \Pi_m$ such that $a_m \leq x \leq b_m$ and $|b_m - a_m| \leq 2^{-m}$. Taking limit as $n\to\infty$ to the inequality $f_n(a_m) \leq f_n(x) \leq f_n(b_m)$, we obtain

  \begin{align*} f(a_m) = \lim_{n\to\infty} f_n (a_m) &\leq \liminf_{n\to\infty} f_n (x) \\ &\leq \limsup_{n\to\infty} f_n (x) \leq \lim_{n\to\infty} f_n (b_m) = f(b_m). \end{align*}

  Taking $m \to \infty$, both $(a_m)$ and $(b_m)$ converge to $x$. Since $x$ is a continuity point of $f$, we have

  $$ f(x) \leq \liminf_{n\to\infty} f_n (x) \leq \limsup_{n\to\infty} f_n (x) \leq f(x) $$

  and hence $f_n(x) \to f(x)$.

Combining altogether, it follows that $f_n \to f$ pointwise on $[0, 1]$ as expected.

Note: This answer dose not answer the OP's question. Just see it as a reference.

Since $f:[0,1]\to \mathbb{R}$ is increasing, it is a bounded Borel measurable function. By Lusin's theorem, there is a sequence $\{f_n\}$ of continuous functions such that $f_n=f$ on a Borel set $E_n$ which satisfies $m\{[0,1]\backslash E_n\}<\frac{1}{2^n}$.

Since $\sum_{n=1}^{\infty}m\{[0,1]\backslash E_n\}<\infty$, for almost every point in $[0,1]$, it is contained in finte sets of $\{[0,1]\backslash E_n\}$ and therefore $f_n$ converges to $f$ almost everywhere.

Here is another answer, which main idea is : if $f$ is right or left continuous, there is an "explicit" solution (see the integrals below).

Claim : Every increasing function $f$ can be written as a sum of an increasing right-continuous function and a left-continuous function.

Proof : the set of points at which $f$ is discontinuous is at most countable : it will be denoted $(a_n)_{n \in \mathbb{N}}$. Now we take $g : x \mapsto \sum \limits_{\substack{n \in \mathbb{N}\\ a_n \le x}} f(a_n)-f(a_n^-)$ and $h = f-g$. It is easy to show that $g$ is right-continuous. If $x \notin \{a_n\}$, $f$ and $g$ are continuous at $x$, so $h$ is continuous at $x$. If $x=a_n$, $g(t) \underset{t \to x^-}{\longrightarrow} g(x)-\big( f(a_n)-f(a_n^-) \big)$, so $h(t) \underset{t \to x^-}{\longrightarrow} f(a_n^-)-g(x)+\big( f(a_n)-f(a_n^-) \big)=h(x)$. Hence $f=g+h$ with $f$ increasing and right-continuous, and $h$ left-continuous.

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If we write $f=g+h$ with $g$, $h$ as in the claim, both $f$ and $g$ are increasing, thus locally bounded and with at most countably many discontinuity points. Hence $f$ and $g$ are Riemann integrable, and so is $h$. For $n \ge 1$, we can thus consider : $$ g_n : x \mapsto n \displaystyle{\int_x^{x+\frac{1}{n}}} g(t)dt,\quad \ h_n : x \mapsto n \displaystyle{\int_{x-\frac{1}{n}}^x} h(t)dt, \quad \ f_n : x \mapsto g_n(x)+h_n(x).$$

As $g$ (resp. $h$) is right (resp. left)-continuous, it is easy to prove that for all $x$, $g_n(x) \underset{n \to +\infty}{\longrightarrow} g(x)$ and $h_n(x) \underset{n \to +\infty}{\longrightarrow} h(x)$, so $\big(f_n\big)_{n \ge 1}$ converges pointwise to $f$. As $g$ and $h$ are locally bounded, $g_n$ and $h_n$ are continuous for all $n$, and thus $\big( f_n \big)_{n \ge 1}$ is a sequence of continuous functions.