As part of some exercise I've been given, we're supposed to identify a function that has the following properties:
- Defined on the interval $ [0,1]$
- Monotonously increasing
- Discontinuous, at every point $x\in\{1/n\}, n=(2,3,...\infty)$
I'm also supposed to show how it is Riemann-integrable. I'm honestly stumped as to how to produce such a function. My first attempt was to define it piecewise as $f(x)=1,x\neq1/n, f(x)=0,x=1/n,$ but that's not a monotonously increasing function.
As far as I've understood with some further reading, you can only have a countable amount of discontinuities for a monotonously increasing function - but I'm tasked with finding one with an infinite amount of discontinuities, so I don't know what I've misunderstood. I can sort of make a picture in my head of a function that jumps up a little at points where $x=1/n$ but I haven't the slightest clue how to define one like it.
For each $n\in\mathbb{N}$, you can define a function $j_n$ by $$ j_n(x) = \begin{cases} 0 & \text{if $x < \frac{1}{n}$, and} \\ 1 & \text{if $x \ge \frac{1}{n}$.} \end{cases} $$ (I am calling these functions $j_n$ because they jump at $\frac{1}{n}$.) Then define the function $f$ by $$ f(x) = \sum_{n=1}^{\infty} \frac{1}{2^n} j_n(x). $$ Observe that $f$ converges pointwise on the interval $[0,1]$. Indeed, with $N = \left\lceil \frac{1}{x} \right\rceil$, we have $$ N = \left\lceil \frac{1}{x} \right\rceil \ge \frac{1}{x} \ge \left\lfloor \frac{1}{x} \right\rfloor > N-1 \implies \frac{1}{N-1} > x \ge \frac{1}{N},$$ from which it follows (modulo a possible off-by-one indexing error, though I'm pretty sure I kept track sufficiently well) that $$ f(x) = \sum_{n=1}^{N} \frac{1}{2^n}, $$ which can be computed explicitly (it is a geometric series). The intuition here is that this is a function that is constant on each interval $\left[ \frac{1}{n+1}, \frac{1}{n} \right)$, then jumps up by $\frac{1}{2^n}$ at $x = \frac{1}{n}$.
It is hopefully not to hard to see that this function is discontinuous at every point of the form $\frac{1}{n}$ (with $n\in\mathbb{N}$, thus there are a countably infinite number of discontinuities), and it is not too hard to show that it is nondecreasing. You can then compute the integral by noting that $$ \int_{0}^{1} f(x) \,\mathrm{d}x = \sum_{n=1}^{\infty} \int_{\frac{1}{n+1}}^{\frac{1}{n}} f(x) \, \mathrm{d}x = \sum_{n=1}^{\infty} \int_{\frac{1}{n+1}}^{\frac{1}{n}} \sum_{k=1}^{n} \frac{1}{2^k} \, \mathrm{d}x. $$ (Exercise: do this explicitly.)
If you need a function that is strictly increasing, consider $$ g(x) = x + f(x), $$ where $f$ is the function defined above.