# 实分析 Real Analysis MAT00005C

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{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
$$\lambda=\int_{a}^{b} f . \diamond$$