 Solution

The printed assertion needs an endpoint qualification: the enumeration includes zero, but $1_{[0,0)}$ is identically zero. In fact $f$ is continuous at zero; the claimed discontinuities occur at every $r_j>0$.

Define the finite step function sumsThe assumed convergence of ${\sum }_{k}1/k^2$ and the bound on each indicator function giveThus the defining series converges everywhere and has uniform convergence. Part (b) proves that $f$ is Riemann integrable; it also gives, if desired,For $r_j>0$ and $0\le t<r_j$, every indicator function is nonincreasing, and the $j$th drops by one between $t$ and $r_j$. HenceTaking inputs from the left tending to $r_j$ proves a discontinuity there.

For completeness, continuity at zero follows directly from a finite-head and tail argument. Given $\epsilon >0$, choose $N$ with ${\sum }_{k>N}1/k^2<\epsilon$. Choose $\delta >0$ below every positive $r_k$ with $k\le N$. For $0\le t<\delta$, the first $N$ indicator functions have exactly their values at zero; the one with $r_k=0$, if present, is always zero. Thus $|f(t)-f(0)|\le {\sum }_{k>N}1/k^2<\epsilon$. This confirms that the endpoint exception is real. Replacing the printed enumeration of $[0,1)$ by one of $(0,1)$ would remove it.

Finally, the density of the rational numbers supplies a positive rational $r_j$ in every nonempty open subinterval of $[0,1]$. Thus a Riemann integrable function can have a discontinuity in every open subinterval: dense jumps do not prevent Riemann integrability.