= Solution
{c}
Let $f:[a,b]\to\mathbb R$ be a bounded <function>, with $a<b$. A <partition of an interval> is $\mathcal P=(a=x_0<x_1<\cdots<x_N=b)$; its <mesh of a partition> is $\|\mathcal P\|=\max_i(x_i-x_{i-1})$. Choose arbitrary tags $\xi_i\in[x_{i-1},x_i]$. The corresponding <Riemann sum> is
$$
\sum(f;\mathcal P,\vec\xi)=\sum_{i=1}^N f(\xi_i)(x_i-x_{i-1}).
$$
The <Riemann sum convergence theorem> says that $f$ is <Riemann integrable> with <Riemann integral> $I$ if and only if, for every $\varepsilon>0$, some $\delta>0$ satisfies
$$
\boxed{\left|\sum(f;\mathcal P,\vec\xi)-I\right|<\varepsilon}
$$
for every <partition of an interval> with $\|\mathcal P\|<\delta$ and every permitted choice of tags. Equivalently, along every <sequence> of such tagged <partitions of an interval> with <mesh of a partition> tending to zero, the <Riemann sums> tend to $I$. Both the shrinking <mesh of a partition> and freedom of tags are essential.
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