Integrability with one wild point

Idea. The oscillation of $f$ near $0$ is large, but it happens on a very short interval, so it costs little area.

Setup. Fix $\eta>0$ and choose $\varepsilon\in(0,1)$ with $\varepsilon<\frac\eta4$.

On $[\varepsilon,1]$. $f$ is continuous, hence integrable, so there is a partition $\sigma'$ of $[\varepsilon,1]$ with $S_{\sigma'}-s_{\sigma'}<\frac\eta2$.

On $[0,\varepsilon]$. Since $-1\le f\le1$, on this single interval $M-m\le2$, contributing at most $2\varepsilon$ to $S-s$.

Total. For the partition $\sigma=\{0\}\cup\sigma'$ of $[0,1]$,
$$S_\sigma-s_\sigma<2\varepsilon+\frac\eta2<\frac\eta2+\frac\eta2=\eta,$$
since $2\varepsilon<\frac\eta2$.
By the Darboux criterion, $f$ is integrable on $[0,1]$.

Remark. The same argument shows that any bounded function with finitely many discontinuities is Riemann integrable. (Compare with the Dirichlet function, discontinuous *everywhere*.)