这是一份YORK约克大学MAT00004C作业代写的成功案例
{Condition (b) is Riemann’s original formulation of integrability. 1 }
(a) $\Rightarrow$ (b): Assuming $f$ is Riemann-integrable, let
$$
\lambda=\int_{a}^{b} f
$$
For each $\nu=1, \ldots, n$, choose a sequence $\left(x_{\nu}^{k}\right)$ in $\left[a_{\nu-1}, a_{\nu}\right]$ such that
$$
f\left(x_{\nu}^{k}\right) \rightarrow M_{\nu} \text { as } k \rightarrow \infty
$$
then
$$
\sum_{\nu=1}^{n} f\left(x_{\nu}^{k}\right) e_{\nu} \rightarrow \sum_{\nu=1}^{n} M_{\nu} e_{\nu}=S(\sigma)
$$
so by $()$ we have (*)
$$
\lambda-\epsilon \leq S(\sigma) \leq \lambda+\epsilon .
$$
MAT00005C COURSE NOTES ：
Similarly,
$$
\lambda-\epsilon \leq s(\sigma) \leq \lambda+\epsilon \text {; }
$$
thus $S(\sigma)$ and $s(\sigma)$ both belong to the interval $[\lambda-\epsilon, \lambda+\epsilon]$, therefore
$$
W_{f}(\sigma)=S(\sigma)-s(\sigma) \leq 2 \epsilon .
$$
This proves that $f$ is Riemann-integrable and since
$$
S(\sigma) \rightarrow \int_{a}^{b} f \text { as } \mathrm{N}(\sigma) \rightarrow 0,
$$
it is clear from $\left({ }^{* *}\right)$ that
$$
\lambda=\int_{a}^{b} f . \diamond
$$
