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# 数学代写|实分析代写Real Analysis代考|Some Riemann integrable functions

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## 数学代写|实分析代写Real Analysis代考|Some Riemann integrable functions

Theorem 11.5,1. Let a functionfs, $[a, b] \rightarrow \mathbb{R}$ be nonatotut $[a, b]$. Then $f$ is intagrable on $[a, b]$.
Proof. Let $f$ be monotone increasing on $[a, b]$. Clearly, $f$ is bounded on $[a, b], f(a)$ being a lower bound and $f(b)$ being an upper bound of $f$ on $[a, b] . f(b)-f(a) \geq 0$
Let us choose $\epsilon>0$.
Let $P$ be a partition of $[a, b]$ with $|P|<\epsilon /{f(b)-f(a)+1}$. Let $P=\left(x_0, x_1, x_2, \ldots, x_n\right)$, where $a=x_0<x_1<x_2<\cdots<x_n=b$.

Let $M_r=\sup {x \in\left[x{r-1}, x_r\right]} f(x), \pi m_r=\inf {x \in\left[x{r-1}, x_r\right]} f(x)$, for $r=1,2, \ldots, n$.
Then $M_r=f\left(x_r\right)$ and $m_r=f\left(x_{r-1}\right)$, for $r=1,2, \ldots, n$.

\text { We have } \begin{aligned} U(P, f)-L(P, f) & =\sum_{r=1}^n\left(M_r-m_r\right)\left(x_r-x_{r-1}\right) \ & =\sum_{r=1}^n\left{f\left(x_r\right)-f\left(x_{r-1}\right)\right}\left(x_r-x_{r-1}\right) \ & \leq|P| \sum_{r=1}^n\left{f\left(x_r\right)-f\left(x_{r-1}\right)\right} \ & =|P|{f(b)-f(a)}<\epsilon . \end{aligned}
Therefore for a chosen positive $\epsilon$, there exists a partition $P$ of $[a, b]$ such that $U(P, f)-L(P, f)<\epsilon$.

This being a suffientandion for integrability, $f$ – is integrable on $[a, b]$.

Proceeding in a similar manner it can be proved that if $f$ be monotone decreasing on $[a, b]$, then $f$ is integrable on $[a, b]$.

## 数学代写|实分析代写Real Analysis代考|Properties of Riemann integrable functions.

Theorem 11 6.1, Lét $f:[a, b] \rightarrow \mathbb{R}, g:[a, b] \rightarrow \mathbb{R}$ be both integrable on $[a, b]$. Then $f+g$ ts integrable on ${a, b} a$ and $f_a^b(f+g)=\int_a^b f_f f_a^b g$.
Proof Since $f \in \mathcal{R}[a, b]$ and $g \in \mathcal{R}[a, b], f$ and $g$ are both bounded on $[a, b]$. Therefore there exist positive real numbers $k_1, k_2$ such that $|f(x)|0$.
Since $f \in \mathcal{R}[a, b]$, there exists a partition $P_1$ of $[a, b]$ such that
$$U\left(P_1, f\right)-L\left(P_1, f\right)<\frac{\epsilon}{2}$$
Since $g \in \mathcal{R}[a, b]$, there exists a partition $P_2$ of $[a, b]$ such that
$$U\left(P_2, g\right)-L\left(P_2, g\right)<\frac{\epsilon}{2}$$
Let $P_0=P_1 \cup P_2$. Then $P_0$ is a refinement of $P_1$ as well as of $P_2$ and $L\left(P_1, f\right) \leq L\left(P_0, f\right) \leq U\left(P_0, f\right) \leq U\left(P_1, f\right)$;
$$L\left(P_2, g\right) \leq L\left(P_0, g\right) \leq U\left(P_0, g\right) \leq U\left(P_2, g\right)$$
So $U\left(P_0, f\right)-L\left(P_0, f\right) \leq U\left(P_1, f\right)-L\left(P_1, f\right)<\frac{\epsilon}{2}$ and $U\left(P_0, g\right)-L\left(P_0, g\right) \leq U\left(P_2, g\right)-L\left(P_2, g\right)<\frac{c}{2}$.
Let $P_0=\left(x_0, x_1, \ldots, x_n\right)$, where $a=x_0<x_1<\cdots<x_n=b$.
Let $M_r=\sup {x \in\left[x{r-1}, x_r\right]}(f+g)(x), m_r=\inf {x \in\left[x{r-1}, x_r\right]}(f+g)(x)$
\begin{aligned} & M_r^{\prime}=\sup {x \in\left[x{r-1}, x_r\right]} f(x), m_r^{\prime}=\inf {x \in\left[x{r-1}, x_r\right]} f(x) \ & M_r^{\prime \prime}=\sup {x \in\left[x{r-1}, x_r\right]} g(x), m_r^{\prime \prime}=\inf {x \in\left[x_r+1, x_r\right]} g(x), \text { for } r=1,2, \ldots, n . \end{aligned} Then $M_r \leq M_r^{\prime}+M_r^{\prime \prime}, m_r \geq m_r^{\prime}+m_r^{\prime \prime}$, for $r=1,2, \ldots, n$. $$\begin{array}{r} U\left(P_0, f+g\right)=M_1\left(x_1-x_0\right)+\cdots+M_n\left(x_n-x{n-1}\right) \ \leq\left[M_1^{\prime}\left(x_1-x_0\right)+\cdots+M_n^{\prime}\left(x_n-x_{n-1}\right)\right] \end{array}$$

\begin{aligned} & +\left[M_1^{\prime \prime}\left(x_1-x_0\right)+\cdots+M_n^{\prime \prime}\left(x_n-x_{n-1}\right)\right] \ & =U\left(P_0, f\right)+U\left(P_0, g\right) . \end{aligned}
Similarly, $L\left(P_0, f+g\right) \geq L\left(P_0, f\right)+L\left(P_0, g\right)$.
Hence $U\left(P_0, f+g\right)-L\left(P_0, f+g\right) \leq\left[U\left(P_0, f\right)-L\left(P_0, f\right)\right]+\left[U\left(P_0, g\right)-\right.$ $\left.L\left(P_0, g\right)\right]<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$.

## 数学代写|实分析代写Real Analysis代考|Some Riemann integrable functions

\text { We have } \begin{aligned} U(P, f)-L(P, f) & =\sum_{r=1}^n\left(M_r-m_r\right)\left(x_r-x_{r-1}\right) \ & =\sum_{r=1}^n\left{f\left(x_r\right)-f\left(x_{r-1}\right)\right}\left(x_r-x_{r-1}\right) \ & \leq|P| \sum_{r=1}^n\left{f\left(x_r\right)-f\left(x_{r-1}\right)\right} \ & =|P|{f(b)-f(a)}<\epsilon . \end{aligned}

$$U\left(P_1, f\right)-L\left(P_1, f\right)<\frac{\epsilon}{2}$$

$$U\left(P_2, g\right)-L\left(P_2, g\right)<\frac{\epsilon}{2}$$

$$L\left(P_2, g\right) \leq L\left(P_0, g\right) \leq U\left(P_0, g\right) \leq U\left(P_2, g\right)$$
$U\left(P_0, f\right)-L\left(P_0, f\right) \leq U\left(P_1, f\right)-L\left(P_1, f\right)<\frac{\epsilon}{2}$和$U\left(P_0, g\right)-L\left(P_0, g\right) \leq U\left(P_2, g\right)-L\left(P_2, g\right)<\frac{c}{2}$。

\begin{aligned} & M_r^{\prime}=\sup {x \in\left[x{r-1}, x_r\right]} f(x), m_r^{\prime}=\inf {x \in\left[x{r-1}, x_r\right]} f(x) \ & M_r^{\prime \prime}=\sup {x \in\left[x{r-1}, x_r\right]} g(x), m_r^{\prime \prime}=\inf {x \in\left[x_r+1, x_r\right]} g(x), \text { for } r=1,2, \ldots, n . \end{aligned}然后是$M_r \leq M_r^{\prime}+M_r^{\prime \prime}, m_r \geq m_r^{\prime}+m_r^{\prime \prime}$，对应$r=1,2, \ldots, n$。 $$\begin{array}{r} U\left(P_0, f+g\right)=M_1\left(x_1-x_0\right)+\cdots+M_n\left(x_n-x{n-1}\right) \ \leq\left[M_1^{\prime}\left(x_1-x_0\right)+\cdots+M_n^{\prime}\left(x_n-x_{n-1}\right)\right] \end{array}$$

\begin{aligned} & +\left[M_1^{\prime \prime}\left(x_1-x_0\right)+\cdots+M_n^{\prime \prime}\left(x_n-x_{n-1}\right)\right] \ & =U\left(P_0, f\right)+U\left(P_0, g\right) . \end{aligned}

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