这是一份anu澳大利亚国立大学MATH3320/MATH6212的成功案例

Suppose that a function $f(x)$ is bounded on the interval $[A, B]$, where $A, B \in \mathbb{R}$ and $A<B$.
DEFINITION. The real number
$$
I^{-}(f, A, B)=\sup {\Delta} s(f, \Delta), $$ where the supremum is taken over all dissections $\Delta$ of $[A, B]$, is called the lower integral of $f(x)$ over $[A, B]$. DEFINITION. The real number $$ I^{+}(f, A, B)=\inf {\Delta} S(f, \Delta),
$$
where the infimum is taken over all dissections $\Delta$ of $[A, B]$, is called the upper integral of $f(x)$ over $[A, B]$.

REMARK. Since $f(x)$ is bounded on $[A, B]$, it follows that $s(f, \Delta)$ and $S(f, \Delta)$ are bounded above and below. This guarantees the existence of $I^{-}(f, A, B)$ and $I^{+}(f, A, B)$.

MATH3320/MATH6212 COURSE NOTES ：

$$
S(f, \Delta)-s(f, \Delta)<\frac{\epsilon}{c} $$ It is easy to see that $$ S(c f, \Delta)=c S(f, \Delta) \quad \text { and } \quad s(c f, \Delta)=c s(f, \Delta) $$ Hence $$ S(c f, \Delta)-s(c f, \Delta)<\epsilon . $$ It follows from Theorem $2 \mathrm{D}$ that $c f \in \mathcal{R}([A, B])$. Also, (13) clearly implies $I^{+}(c f, A, B)=c I^{+}(f, A, B)$. Suppose next that $c<0$. Since $f \in \mathcal{R}([A, B])$, it follows from Theorem $2 \mathrm{D}$ that for every $\epsilon>0$, there exists a dissection $\Delta$ of $[A, B]$ such that
$$
S(f, \Delta)-s(f, \Delta)<-\frac{\epsilon}{c}
$$
It is easy to see that
$$
S(c f, \Delta)=c s(f, \Delta) \quad \text { and } \quad s(c f, \Delta)=c S(f, \Delta)
$$
Hence
$$
S(c f, \Delta)-s(c f, \Delta)<\epsilon .
$$
It follows from Theorem $2 \mathrm{D}$ that $c f \in \mathcal{R}([A, B])$. Also, (14) clearly implies $I^{+}(c f, A, B)=c I^{-}(f, A, B)$.
(c) Note simply that
$$
\int_{A}^{B} f(x) \mathrm{d} x \geq(B-A) \inf _{x \in[A, B]} f(x)
$$
where the right hand side is the lower sum corresponding to the trivial dissection.
(d) Note that $g-f \in \mathcal{R}([A, B])$ in view of (a) and (b). We now apply (c) to the function $g-f$.
Next, we investigate the question of breaking up the interval $[A, B]$ of integration.