Lemma
$A, B \subseteq \mathbb{R}$, $\lambda \in \mathbb{R}$
1. $\mathrm{sup}(A + B) = \mathrm{sup}A + \mathrm{sup}B$
2. $\mathrm{inf}(A + B) = \mathrm{inf}A + \mathrm{inf}B$
3. $\mathrm{sup}(\lambda A) = \begin{cases}
\lambda \; \mathrm{sup} A & \text{ if } \; \lambda \geq 0 \\
\lambda \; \mathrm{inf} A & \text{ if } \; \lambda < 0
\end{cases}$
4. $\mathrm{inf}(\lambda A) = \begin{cases}
\lambda \; \mathrm{inf} A & \text{ if } \; \lambda \geq 0 \\
\lambda \; \mathrm{sup} A & \text{ if } \; \lambda < 0
\end{cases}$
정리1
$f, g \in \mathcal{R}[a, b]$
$\Rightarrow$ $f + g \in \mathcal{R}[a, b]$ & $\int_{a}^{b} f + g dx = \int_{a}^{b} f dx + \int_{a}^{b} g dx$
Claim 1 : 임의의 분할 $P \in \mathcal{P}[a, b]$에 대하여
$U(f+g, P) \leq U(f,P) + U(g,P)$ & $L(f+g, P) \geq L(f,P) + L(g, P)$
임의의 분할 $P = \left\{x_{0}, x_{1}, \cdots, x_{n} \right\} \in \mathcal{P}[a, b]$를 택하자.
각 소구간 $I_{k} = [x_{k-1}, x_{k}]$에 대하여
$\left\{f(x)+g(x) \; | \; x \in I_{k} \right\} \subseteq \left\{f(x) \; | \; x \in I_{k} \right\} + \left\{g(x) \; | \; x \in I_{k} \right\}$이므로
$U(f+g, P) = \sum_{k=1}^{n} \mathrm{sup} \left\{f(x)+g(x) \; | \; x \in I_{k} \right\} \Delta x_{k}$
$\leq \sum_{k=1}^{n} \mathrm{sup} \left\{f(x) \; | \; x \in I_{k} \right\} \Delta x_{k} + \sum_{k=1}^{n} \mathrm{sup} \left\{g(x) \; | \; x \in I_{k} \right\} \Delta x_{k}$
$=U(f,P) + U(g,P)$
비슷한 방법으로 $L(f+g, P) \geq L(f,P) + L(g, P)$
Show 1 : $f + g \in \mathcal{R}[a, b]$
임의의 $\varepsilon > 0$을 택하자.
$f, g \in \mathcal{R}[a, b]$이므로 적당한 $P_{1}, P_{2} \in \mathcal{P}[a, b]$가 존재하여
$ U(f, P_{1}) - L(f, P_{1}) < \frac{\varepsilon}{2}$
$U(g, P_{2}) - L(g, P_{2}) < \frac{\varepsilon}{2}$
를 만족한다.
$P_{\varepsilon} = P_{1} \cup P_{2}$라 하면
$U(f, P_{\varepsilon}) - L(f, P_{\varepsilon}) \leq U(f, P_{1}) - L(f, P_{1}) < \frac{\varepsilon}{2} $
$U(g, P_{\varepsilon}) - L(g, P_{\varepsilon}) \leq U(g, P_{2}) - L(g, P_{2}) < \frac{\varepsilon}{2} $
이다. 즉,
$U(f+g, P_{\varepsilon}) - L(f+g, P_{\varepsilon}) \leq U(f, P_{\varepsilon}) + U(g, P_{\varepsilon}) - L(f, P_{\varepsilon}) - L(g, P_{\varepsilon})$
$= (U(f, P_{\varepsilon}) - L(f, P_{\varepsilon})) + (U(g, P_{\varepsilon}) - L(g, P_{\varepsilon}))$
$\leq (U(f, P_{1}) - L(f, P_{1})) - (U(g, P_{2}) - L(g, P_{2}))$
$< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$
$\therefore$ $f + g \in \mathcal{R}[a, b]$
Show 2 : $\int_{a}^{b} f + g dx = \int_{a}^{b} f dx + \int_{a}^{b} g dx$
이때
$\int_{a}^{b} f + g dx \leq U(f+g, P_{\varepsilon })$
$\leq U(f, P_{\varepsilon }) + U(g, P_{\varepsilon })$
$< L(f, P_{\varepsilon }) + L(g, P_{\varepsilon }) + \varepsilon $
$\leq \int_{a}^{b} f dx + \int_{a}^{b} g dx + \varepsilon $
$\Rightarrow$ $ \int_{a}^{b} f + g dx \leq \int_{a}^{b} f dx + \int_{a}^{b} g dx$
또한
$\int_{a}^{b} f dx + \int_{a}^{b} g dx \leq U(f, P_{\varepsilon }) + U(g, P_{\varepsilon })$
$< L(f, P_{\varepsilon }) + L(g, P_{\varepsilon }) + \varepsilon $
$\leq L(f+g, P_{\varepsilon }) + \varepsilon $
$\leq \int_{a}^{b} f + g dx + \varepsilon$
$\Rightarrow$ $\int_{a}^{b} f dx + \int_{a}^{b} g dx \leq \int_{a}^{b} f + g dx $
$\therefore$ $\int_{a}^{b} f + g dx = \int_{a}^{b} f dx + \int_{a}^{b} g dx$
정리2
$f, g \in \mathcal{R}[a, b]$, $\lambda \in \mathbb{R}$
$\Rightarrow$ $\lambda f \in \mathcal{R}[a, b]$ & $\int_{a}^{b} \lambda f dx = \lambda \int_{a}^{b} f dx $
($\lambda >0$인 경우만)
임의의 분할 $P \in \mathcal{P}[a, b]$를 택하자.
$U(\lambda f, P) = \sum_{k=1}^{n} \mathrm{sup} \left\{\lambda f(x) \; | \; x \in I_{k} \right\} \Delta x_{k}$
$ = \sum_{k=1}^{n} \lambda \; \mathrm{sup} \left\{ f(x) \; | \; x \in I_{k} \right\} \Delta x_{k}$
$ = \lambda \; U(f, P)$
$\therefore$ $\overline{\int_{a}^{b}} \lambda f dx = \mathrm{inf} \left\{U(\lambda f, P) \; | \; P \in \mathcal{P}[a, b]\right\}$
$= \lambda \; \mathrm{inf} \left\{U(f, P) \; | \; P \in \mathcal{P}[a, b]\right\}$
$= \lambda \overline{\int_{a}^{b}} f dx$
$= \lambda \int_{a}^{b} f dx$
비슷한 방법으로 $\underline{\int_{a}^{b}} \lambda f dx = \lambda \int_{a}^{b} f dx $
$\therefore$ $\lambda f \in \mathcal{R}[a, b]$ & $\int_{a}^{b} \lambda f dx = \lambda \int_{a}^{b} f dx $
# 나머지도 비슷한 방법으로 증명하면 된다.
정의1
$f^{+}(x) = \mathrm{max} \left\{f(x), 0 \right\}$, $f^{-}(x) = - \mathrm{min} \left\{f(x), 0 \right\} $로 정의하자.
# $f(x) = f^{+}(x) - f^{-}(x)$, $\left|f(x) \right| = f^{+}(x) + f^{-}(x)$
정리3
$f \in \mathcal{R}[a, b]$이면
1. $f^{+}, f^{-} \in \mathcal{R}[a, b]$
2. $\left|f(x) \right| \in \mathcal{R}[a, b]$ & $\left|\int_{a}^{b} f dx \right| \leq \int_{a}^{b} \left|f \right| dx$
pf)
임의의 분할 $P \in \mathcal{P}[a, b]$에 대하여
$M(f, I_{k}) - m(f, I_{k}) \geq M(f^{+}, I_{k}) - m(f^{+}, I_{k}) $
를 만족한다.
또한 $f \in \mathcal{R}[a, b]$이므로 임의의 $\varepsilon > 0$에 대하여 적당한 $P_{\varepsilon} \in \mathcal{P}[a, b]$가 존재하여
$U(f, P_{\varepsilon}) - L(f, P_{\varepsilon}) < \varepsilon$
를 만족한다.
$\therefore$ $U(f^{+}, P_{\varepsilon}) - L(f^{+}, P_{\varepsilon}) $ $= \sum_{k=1}^{n} (M(f^{+}, I_{k}) - m(f^{+}, I_{k})) \Delta x_{k}$
$\leq \sum_{k=1}^{n} (M(f, I_{k}) - m(f, I_{k})) \Delta x_{k}$
$= U(f, P_{\varepsilon}) - L(f, P_{\varepsilon}) < \varepsilon$
$\therefore$ $f^{+} \in \mathcal{R}[a, b]$ & $f^{-} = f^{+} - f \in \mathcal{R}[a, b]$
$\left|f \right| = f^{+} + f^{-} \in \mathcal{R}[a, b]$
$\left|\int_{a}^{b} f dx \right| = \left|\int_{a}^{b} f^{+} dx - \int_{a}^{b} f^{-} dx \right|$
$\leq \int_{a}^{b} f^{+} dx + \int_{a}^{b} f^{-} dx$
$= \int_{a}^{b} \left|f \right| dx$
정리4
$f, g \in \mathcal{R}[a, b]$
$\Rightarrow$ $fg \in \mathcal{R}[a, b]$
# 증명생략
정리5
$f : [a, b] \to \mathbb{R}$ 유계, $c \in (a,b)$에 대하여
$f \in \mathcal{R}[a, b]$ $\Leftrightarrow$ $f \in \mathcal{R}[a, c]$ & $f \in \mathcal{R}[c, b]$
이 경우에 $\int_{a}^{b} f dx = \int_{a}^{c} f dx + \int_{c}^{b} f dx$
pf)
$\Rightarrow)$
Show 1 : $f \in \mathcal{R}[a, c]$ & $f \in \mathcal{R}[c, b]$
임의의 $\varepsilon > 0$을 택하자.
$f \in \mathcal{R}[a, b]$이므로 적당한 $P_{\varepsilon}$이 존재하여
$U(f, P_{\varepsilon}) - L(f, P_{\varepsilon}) < \varepsilon$
를 만족한다.
$P' = P_{\varepsilon} \cup \left\{c \right\}$라 하면
$ U(f, P') - L(f, P') \leq U(f, P_{\varepsilon}) - L(f, P_{\varepsilon}) < \varepsilon$
이때 $P_{1} = P' \cap [a, c] $라 하면 $P_{1} \in \mathcal{P}[a, c]$이고
$U(f, P_{1}) - L(f, P_{1}) \leq U(f, P') - L(f, P') < \varepsilon $
$\therefore$ $f \in \mathcal{R}[a, c]$
또한 $P_{2} = P' \cap [c, b] $라 하면 $P_{2} \in \mathcal{P}[c, b]$이고
$U(f, P_{2}) - L(f, P_{2}) \leq U(f, P') - L(f, P') < \varepsilon $
$\therefore$ $f \in \mathcal{R}[c, b]$
Show 2 : $\int_{a}^{b} f dx = \int_{a}^{c} f dx + \int_{c}^{b} f dx$
이때
$\int_{a}^{b} f dx \leq U(f, P_{\varepsilon })$
$< L(f, P_{\varepsilon }) + \varepsilon $
$= L(f, P_{1}) + L(f, P_{2}) +\varepsilon $
$\leq \int_{a}^{c} f dx + \int_{c}^{b} f dx + \varepsilon $
$\Rightarrow$ $\int_{a}^{b} f dx \leq \int_{a}^{c} f dx + \int_{c}^{b} f dx$
또한
$\int_{a}^{c} f dx + \int_{c}^{b} f dx \leq U(f, P_{1}) + U(f, P_{2})$
$< L(f, P_{1}) + L(f, P_{2}) + \varepsilon$
$= L(f, P_{\varepsilon }) + \varepsilon $
$\leq \int_{a}^{b} f dx + \varepsilon $
$\Rightarrow$ $\int_{a}^{c} f dx + \int_{c}^{b} f dx \leq \int_{a}^{b} f dx$
$\therefore$ $\int_{a}^{b} f dx = \int_{a}^{c} f dx + \int_{c}^{b} f dx$
$\Leftarrow)$
Show 1 : $f \in \mathcal{R}[a, b]$
임의의 $\varepsilon > 0$을 택하자.
$f \in \mathcal{R}[a, c]$, $f \in \mathcal{R}[c, b]$이므로 적당한 $P_{1} \in \mathcal{P}[a, c]$, $P_{2} \in \mathcal{P}[c, b]$ 이 존재하여
$U(f, P_{1}) - L(f, P_{1}) < \frac{\varepsilon}{2}$
$U(f, P_{2}) - L(f, P_{2}) < \frac{\varepsilon}{2}$
를 만족한다.
$P_{\varepsilon} = P_{1} \cup P_{2}$라 하면 $P_{\varepsilon} \in \mathcal{P}[a, b]$이고
$U(f, P_{\varepsilon}) - L(f, P_{\varepsilon}) = U(f, P_{1}) - U(f, P_{2}) - L(f, P_{1}) - L(f, P_{2}) < \varepsilon $
$\therefore$ $f \in \mathcal{R}[a, b]$
Show 2 : $\int_{a}^{b} f dx = \int_{a}^{c} f dx + \int_{c}^{b} f dx$
위와 동일하다.