Is the function $f_n= \chi_{[\frac{j}{2^k} , \frac{j+1}{2^k}]}$, where $n=2^k+j$ with $0 \leq 2^k$, continuous on the interval $[0,1]$? I have no intuition regarding how this function looks geometrically and also why we have that $\int^{1}_{0}|f_n|=2^{-k}$ for $2^k\leq n<2^{k+1}$.
Do you have any tips as to how to think about this kind of function, especially how to integrate this kind of function with the Riemann Integral, if possible? Any feedback is greatly appreciated.
On
Some related graphs, in the name of 'geometric intuition'-
(Interactive graph with adjustable parameters $j,k$)
The top two graphs are continuous, the bottom two graphs correspond to the function in your question. Essentially, the function in the question is the result of rescaling and translating the simpler function $\chi_{[0,1]}$. The rescaling has a linear effect on the area, while translation has no effect.
We have
$f_n(x)=1$ for $x \in [\frac{j}{2^k} , \frac{j+1}{2^k}]$ and $f_n(x)=0$ if $x \in [0,1] \setminus [\frac{j}{2^k} , \frac{j+1}{2^k}]$.
$f_n$ is not continuous at $\frac{j}{2^k}$ and at $\frac{j+1}{2^k}$
Furthermore $\int^{1}_{0}|f_n| dx=\int^{1}_{0}f_n dx=\int^{\frac{j+1}{2^k}}_{\frac{j}{2^k}}1 dx=\frac{j+1}{2^k}-\frac{j}{2^k}=\frac{1}{2^k}$.