How to construct Riemann Integrable functions on [0,1], that have dense set of discontinuities.p.1 One can  construct Riemann Integrable functions on [0,1], that have a dense set of discontinuities as follows.
(a) Let $f(x) =0$ when $x < 0$, and $f(x) = 1$ if $x\geq$ 0. Choose a countable dense sequence ${r_n}$ in [0,1]. Then Show that the function $$F(x) = \sum_{n = 1}^{\infty}\frac{1}{n^2}f(x - r_n)$$.
is integrable and has discontinuities at all the points of the sequence ${r_n}$.Do I have to show that $F$ is monotonic and bounded? and the question has parts(b) & (c) also, Do part (a) is a way to construct Riemann Integrable functions on [0,1], that have a dense set of discontinuities or part (a) combined with part (b) & (c)?
(b)Consider next$$F(x) = \sum_{n=1}^{\infty}3^{-n} g(x- r_{n})$$.
where $g(x)= sin\frac{1}{x}$ when  $x\ne 0$, and $g(0) = 0$. Then $f$ is integrable, discontinous at each  $x = r_n$, and fails to be monotonic in any subinterval of [0,1].
Could anyone help me?  
 A: I will give an outline of a proof here.
For (a), it is easy to show that $F$ is monotonic and bounded. It is monotonic because each summand is. It is bounded because $F(1) = \sum_{n=1}^\infty 1/n^2$ converges. (We should also show that $F(x)$ converges everywhere on [0,1] because it absolutely converges.)
The rest is to use the following theorem, whose proof can be found on Page 284 (Proposition 1.3).
A bounded monotonic function f on an interval [a, b] is integrable.
For (b), $F(x)$ again absolutely converges, and by proving this, you show at the same time that $F(x)$ is bounded.
To show it is not monotonic, see this answer.
To show it is Riemann integrable, you need to show that $F(x)$ is continuous everywhere except at points in {$r_n$}.
For any $x_0 \not\in \{r_n\} $ and any $\epsilon > 0$, choose an $N$ such that $3^{-N} < \epsilon$, and then choose an interval $(a',b')$ around $x_0$ so that $r_k \not\in (a',b')$ for $k \leq N$.
We then choose an interval $(a,b)$ around $x_0$, such that

*

*$(a,b) \subset (a',b')$, and

*the total fluctuation inside $(a,b)$ caused by $3^{-n} g(x-x_n)$ for $n\leq N$ is less than $\epsilon$. This can be done because there is a finite number of them.

For any $x \in (a,b)$, we have $F(x) - F(x_0) \leq \epsilon + 2 \sum_{n>N} 3^{-n} < \epsilon + 2 \cdot 3^{-N} < 3\epsilon$.
This proves that $F(x)$ is continuous at $x_0$.
Now simply use the following theorem on Page 287 (Theorem 1.7).
A bounded function $f$ on $[a, b]$ is integrable if and only if its set of discontinuities has measure 0.
