= Solution
For a bounded function, a <partition of an interval> is $P:0=x_0<x_1<\cdots<x_N=1$. Put $m_j=\inf_{[x_{j-1},x_j]}f$ and $M_j=\sup_{[x_{j-1},x_j]}f$. Its <lower Darboux sum> and <upper Darboux sum> are
$$
L(f,P)=\sum_{j=1}^Nm_j(x_j-x_{j-1}),\qquad
U(f,P)=\sum_{j=1}^NM_j(x_j-x_{j-1}).
$$
The <lower Darboux integral> is $\sup_P L(f,P)$ and the <upper Darboux integral> is $\inf_P U(f,P)$. \b[<Riemann integrability> means that these two real numbers coincide], and their common value is the <Riemann integral>.
Equivalently, the <Riemann integrability criterion> says that for every $\varepsilon>0$ some partition satisfies $U(f,P)-L(f,P)<\varepsilon$. To see equivalence, all lower sums are at most all upper sums: refine any two partitions to their common refinement and use <Darboux sum refinement monotonicity>. Thus a small gap forces the difference of the two integrals below $\varepsilon$. Conversely, if both integrals equal $I$, choose a lower sum above $I-\varepsilon/2$ and an upper sum below $I+\varepsilon/2$; their common refinement has gap below $\varepsilon$.
This also agrees with the tagged <Riemann sum> definition: there is $I$ such that every tagged sum on a partition of sufficiently small mesh is within any prescribed tolerance of $I$. Each tagged sum lies between the two <Darboux sums>. From a partition with small gap, an arbitrary small-mesh partition differs from its refinement at only the cells crossing the finitely many original division points; boundedness makes their total contribution tend to zero. Conversely, on a fine partition tags approximating each cell <supremum> or <infimum> force both <Darboux sums> close to the same tagged-sum <limit>.
Now suppose $f$ is continuous on $[0,1]$. It is bounded by the result of Question 11. It is also uniformly continuous: otherwise there are $x_n,y_n\in[0,1]$ with $|x_n-y_n|\to0$ but $|f(x_n)-f(y_n)|\geq\varepsilon_0>0$. The <Bolzano-Weierstrass theorem> gives a convergent subsequence of $x_n$; the corresponding $y_n$ have the same <limit>, contradicting <continuity>. Given $\varepsilon>0$, this <uniform continuity> gives $\delta>0$ such that points less than $\delta$ apart have function values less than $\varepsilon/2$ apart. A partition of mesh below $\delta$ then satisfies
$$
U(f,P)-L(f,P)\leq\frac{\varepsilon}2\sum_j(x_j-x_{j-1})=\frac{\varepsilon}2<\varepsilon.
$$
Hence \b[every <continuous function> on the interval is Riemann integrable], with the required proof supplied by its oscillation estimates.
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