Thomae's function is nonnegative, Riemann integrable on $[0,1]$ with integral $0$, and nonzero at every rational, so a vanishing integral does not force a nonnegative integrand to vanish

Refuted: that a nonnegative Riemann integrable function on $[a,b]$ with ${\int }_a^{b}f=0$ vanishes identically (FALSE: a nonnegative Riemann integrable function on $[a,b]$ with ${\int }_a^{b}f=0$ is identically zero, The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as ${\sup }_{P}L(f,P)$ and ${\inf }_{P}U(f,P)$, Darboux integrability as their equality, and the notation ${\int }_a^{b}f$).

The witness is Thomae's function $t$ on $[0,1]$ (The Dirichlet function $1_{\mathbb{Q}}$, and Thomae's function $t$ with $t(x)=1\mathrm{/}q$ at a rational $x=p\mathrm{/}q$ in lowest terms with $q\ge 1$ and $t(x)=0$ at every irrational $x$). It satisfies $0\le t\le 1$, it is Riemann integrable with ${\int }_0^{1}t=0$ (Thomae's function is Riemann integrable on $[0,1]$ with integral $0$: it is continuous at every irrational, so its discontinuity set is countable, and every lower Darboux sum is $0$), and it is positive at every rational point of $[0,1]$ — a dense set, and an infinite one. So the failure is not at a single stray point: the function is nonzero on a dense subset of the interval and the integral still vanishes.

The repaired statement asks for continuity. With $f$ continuous the conclusion is true, and the reason is exactly what fails here: a continuous function positive at one point is positive on a whole subinterval, whereas $t$ is positive only on a set that contains no interval (Both $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable). That repaired statement is not proved here, since the additivity of the integral over subintervals is not available at this point in the reading order.

• What the integral measures, and what it does not. ${\int }_0^{1}t=0$ says that the upper Darboux sums can be made arbitrarily small, not that $t$ is small anywhere in particular. The set where $t$ is positive is $\mathbb{Q}\cap [0,1]$, which is null (Every at most countable subset of $\mathbb{R}$ has measure zero); by Lebesgue's criterion for Riemann integrability: a bounded $f$ on $[a,b]$ is Riemann integrable if and only if its set of discontinuities has measure zero that is also exactly why $t$ is integrable at all.

• Every upper Darboux sum is strictly positive. Each subinterval contains a rational, so $M_i>0$ for every $i$ and $U(t,P)>0$ for every partition $P$; the upper integral is nevertheless $0$, an infimum of a set of positive numbers. Nothing is contradictory here, and it is the cleanest reminder that an infimum need not be attained.

• The Dirichlet function is not available as a witness. It is also nonnegative and nonzero on a dense set, but it is not integrable at all (The Dirichlet function on $[0,1]$ has lower Darboux integral $0$ and upper Darboux integral $1$, so it is bounded and not Riemann integrable), so it cannot satisfy the hypotheses of the refuted claim. Thomae's function is the standard witness precisely because it repairs integrability while keeping the dense positive set.