= Solution
{c}
\b[The printed assertion needs an endpoint qualification:] the enumeration includes zero, but $\mathbf1_{[0,0)}$ is identically zero. In fact $f$ is continuous at zero; the claimed <discontinuities> occur at every $r_j>0$.
Define the finite <step function> sums
$$
\phi_N(t)=\sum_{k=1}^N\frac1{k^2}\mathbf1_{[0,r_k)}(t).
$$
The assumed convergence of $\sum_k1/k^2$ and the bound on each <indicator function> give
$$
0\leq f(t)-\phi_N(t)\leq\sum_{k>N}\frac1{k^2},\qquad\|f-\phi_N\|_\infty\leq\sum_{k>N}\frac1{k^2}\longrightarrow0.
$$
Thus the defining <series> converges everywhere and has <uniform convergence>. Part (b) proves that \b[$f$ is <Riemann integrable>]; it also gives, if desired,
$$
\boxed{\int_0^1f(t)\,dt=\sum_{k=1}^{\infty}\frac{r_k}{k^2}.}
$$
For $r_j>0$ and $0\leq t<r_j$, every <indicator function> is <nonincreasing>, and the $j$th drops by one between $t$ and $r_j$. Hence
$$
f(t)-f(r_j)=\sum_{k\geq1}\frac{\mathbf1_{[0,r_k)}(t)-\mathbf1_{[0,r_k)}(r_j)}{k^2}\geq\frac1{j^2}.
$$
Taking 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 $\varepsilon>0$, choose $N$ with $\sum_{k>N}1/k^2<\varepsilon$. Choose $\delta>0$ below every positive $r_k$ with $k\leq N$. For $0\leq t<\delta$, the first $N$ <indicator functions> have exactly their values at zero; the one with $r_k=0$, if present, is always zero. Thus $|f(t)-f(0)|\leq\sum_{k>N}1/k^2<\varepsilon$. 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 \b[a <Riemann integrable> <function> can have a <discontinuity> in every open subinterval]: <dense jumps do not prevent Riemann integrability>.
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