Oscillation of a function at a point We define $$Osc_f(x):=\inf_{r>0}Osc_f\left((x-r,x+r)\right)$$
Where,$$Osc_f(U)=\sup_{x\in U}f(x)-\inf_{x\in U}f(x)$$
Now we define $f:\mathbb{R}\to[0,1)$ on binary digits as follows,
$$f(x)=\sum_{k=1}^{\infty}\frac{\beta_{2k}(x)}{2^k}$$ for $x=\sum_{k=1}^{\infty}\frac{\beta_k(x)}{2^k}$ where $\beta_k(x)\in \{0,1\}$ and $\liminf_k\beta_k(x)=0$. $f(x)=0 \text{ for } x\in \mathbb{R}\setminus(0,1)$.
I need to prove that $Ocs_f(x)=2^{-m}$ if $x=\frac{2k+1}{2^{2m+1}}$ where $m\in\mathbb{N}$ and
$0\leq k\le2^{2m}-1$.
If it helps, the graph of $f(x)$,

I have no I idea or a lead. Would love some advise.
 A: If $x = \dfrac{2k + 1}{2^{2m + 1}}$ then the binary representation of $x$ consists of binary representation of $k$ till $2m$ binary digits and then followed by $1$. This means that if $$k = b_{1}b_{2}\ldots b_{2m}$$ in binary notation then $$x = 0.b_{1}b_{2}\ldots b_{2m}1$$ in the binary notation. It follows that $\beta_{i}(x) = b_{i}$ for $i = 1, 2, \ldots 2m$ and $\beta_{2m + 1}(x) = 1$.
Hence $$f(x) = \sum_{n = 1}^{\infty}\frac{\beta_{2n}(x)}{2^{n}} = \sum_{n = 1}^{m}\frac{b_{2n}}{2^{n}}$$ Now we can choose a neighborhood $I$ of $x$ so small that the first $2m$ binary digits of any number in the neighborhood $I$ are same as that of $x$. Let $y$ be any such point in this neighborhood $I$ then $$y = 0.b_{1}b_{2}\ldots b_{2m}c_{1}c_{2}\ldots$$ and then we can see that $$f(y) = \sum_{n = 1}^{m}\frac{b_{2n}}{2^{n}} + \sum_{n = 1}^{\infty}\frac{c_{2n}}{2^{m + n}} = f(x) + \sum_{n = 1}^{\infty}\frac{c_{2n}}{2^{m + n}}$$ so it follows that infimum of $f$ occurs when all the $c_{i}$ are $0$ and therefore $$\inf_{y \in I}f(y) = f(x)$$ and clearly supremum occurs when all the $c_{i}$ are $1$ and hence $$\sup_{y \in I}f(y) = f(x) + \sum_{n = 1}^{\infty}\frac{1}{2^{m + n}} = f(x) + \frac{1}{2^{m}}$$ Note that for any number $y \in I$ with $y < x$ we have $c_{1} = 0$ and for any number $y \in I$ with $y > x$ we must have $c_{1} = 1$. Clearly none of this impacts $f(y)$ because its value is dependent on $c_{2}, c_{4}$ etc.
It follows that $$\text{Osc}_{f}(I) = \frac{1}{2^{m}}$$ however small neighborhood $I$ we choose around $x$. Thus $\text{Osc}_{f}(x) = 2^{-m}$.

You should also be able to prove that for all other points in $(0, 1)$ the oscillation of $f$ is $0$. Thus $f$ is continuous in $(0, 1)$ except for a countable number of points in given by $x = (2k + 1)/2^{2m + 1}$. Further one should note that even if $f$ is discontinuous at these exceptional points it is possible to choose value of $m$ such that the oscillation of $f$ at $x$ i.e. $1/2^{m}$ can be made smaller than any given quantity.
