= Solution
Let $L$ be the supremum of all <lower Darboux sums> and $U$ the infimum of all <upper Darboux sums>. The assumed comparison of arbitrary lower and upper sums gives $L\leq U$. For the partition supplied for a given $\varepsilon$,
$$
0\leq U-L\leq S_{\mathcal D}(f)-s_{\mathcal D}(f)<\varepsilon.
$$
Since this holds for every positive $\varepsilon$, $U=L$. Thus the <Riemann integrability criterion> gives
$$
\boxed{f\text{ is Riemann integrable}.}
$$
If $f$ is continuous on $[a,b]$, the <Heine-Cantor theorem> makes it uniformly continuous. Given $\varepsilon>0$, choose $\delta$ so that intervals of length below $\delta$ have oscillation below $\varepsilon/(b-a)$. A partition of mesh below $\delta$ then satisfies
$$
S_{\mathcal D}(f)-s_{\mathcal D}(f)<\varepsilon,
$$
proving that <Continuous functions are Riemann integrable>.
For the function defined using $g$, choose $K$ with $|f|\leq K$. Given $\varepsilon>0$, choose $\delta>0$ so small that $2K\delta<\varepsilon/2$. On $[\delta,1]$, the function $g$ is continuous and therefore has a partition whose upper-minus-lower sum is below $\varepsilon/2$. Adding the interval $[0,\delta]$ contributes at most $2K\delta$, regardless of the value $f(0)=\lambda$. Hence the full Darboux-sum difference is below $\varepsilon$, and
$$
\boxed{f\text{ is Riemann integrable for every }\lambda\in\mathbb R}.
$$
Finally, for any partition $a=x_0<\cdots<x_n=b$, the <mean value theorem> supplies $\xi_i\in(x_{i-1},x_i)$ such that
$$
f(x_i)-f(x_{i-1})=f'(\xi_i)(x_i-x_{i-1}).
$$
Summing telescopes to
$$
f(b)-f(a)=\sum_{i=1}^nf'(\xi_i)(x_i-x_{i-1}),
$$
a <Riemann sum> for the Riemann-integrable function $f'$. Equivalently, every lower derivative sum is at most this telescoping value and every upper derivative sum is at least it. Taking the common <upper and lower Darboux integrals> proves the <fundamental theorem of calculus> conclusion
$$
\boxed{\int_a^bf'(x)\,dx=f(b)-f(a)}.
$$
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