Prove that there exists a sequence of continuous functions $f_n(x)$ such that $f_n \rightarrow f$ pointwise on this interval. Suppose that the real-valued function $f(x)$ is nondecreasing on the interval $[0,1]$. Prove that there exists a sequence of continuous functions $f_n(x)$ such that $f_n \rightarrow f$ pointwise on this interval.
I've been considering problem for a while. The main results that I know of are all about converging almost everywhere. I want to construct $f_n$ but don't know where to start.
 A: since the function is increasing there is only a finite number of jumps of bigger than size $1/n$ . So I'd define an approximating function using this fact.
fix $n$. remove all jumps of size less than $1/n.$ We now have a function cts except at a finite number of points. Make the function grow linearly from left limits to right limit on an interval of size $1/n^2$ at each of these and keep it continuous elsewhere. 
Something like this should converge to what you want. 
OK there is still a lot of work to do but it's a place to start.
A: The problem is easy enough if $f(x)$ is also bounded:
Let $$f_n(x) = f\left( 2^{-n} \left\lfloor 2^n x \right\rfloor \right)
+\left( x- 2^{-n} \left\lfloor 2^n x \right\rfloor \right)
f\left( 2^{-n} \left\lfloor 2^n x \right\rfloor +2^{-n}\right)
$$
One can picture each $f_n$ as a continuous function which is linear in each of the $2^n$ divisions of the unit interval, and matches $(f(x)$ at the endpoints of each division; it is what illustrations of the trapezoid rule look like. 
Since $f(x)$ is bounded, no interval contains a change of more than $M \equiv f(1)-f(0)$, so 
$$|f_n(x) - f(x)| \leq 2^{-n} M$$ which makes it easy to explicitly find an $N(\epsilon)$ for any given $\epsilon$ such that  $$n \geq N \implies |f_n(x) - f(x)| < \epsilon$$
that is, the sequence $\{f_n\} \to f$ pointwise.
If the interval specified were open, e.g., $(0,1)$, but the restriction that $f(x)$ is bounded is removed, the same sort of constructive proof applies.  Here, the $f_n(x)$ are only defined as above on $[2^{-n},1-2^{-n}]$ and outside that interval we can let $f_n(x)$ be $f_n(2^{-n})$ on the left and $f_n(1-2^{-n})$ on the right.  But the pointwise approach of $f_n$ to $f$ still holds, since for any given point $x\in (0,1)$ we can impose a further condition on $N(\epsilon)$ so that $n$ must be large enough that $x$ is in the "good" region for $f_n$.
The above construction works, for example, on the pretty nasty function $f(x) = -\left\lfloor\frac{1}{x}\right\rfloor$
We are left with the case of an unbounded non-decreasing function on $[0,1]$, and of course, that does not exist in standard real analysis.
