if $f:[0,1] \to \mathbb{R}$ is increasing, show that $f$ is the pointwise limit of a sequence of continuous functions over $[0,1]$ if $f:[0,1] \to \mathbb{R}$ is increasing, show that $f$ is the pointwise limit of a sequence of continuous  functions over $[0,1]$
Intuitively this makes sense but I am having trouble with showing why there would be a sequence of continuous functions converging pointwise to $f$.
Clearly there is a sequence converging pointwise to $f$, I can set:
$\forall n \in \mathbb{N}, f_n = f$.
How to prove there is at least one which is made up of continuous functions $f_n, \forall n \in \mathbb{N}$ over $[0,1]$ I can't quite figure out the argument. 
 A: 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.
A: 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.
A: 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.
$ $
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.
A: So: monotone functions are Baire-one functions, as you can see here and here. This means, by definition, that elements of this class are pointwise limits of continuous functions, see here. Hence the statement is proved.
