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

# 实分析 Real Analysis MATH20101

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conclusion holds for $k=1$. Let us proceed by induction on $k$. If $C$ has no interior, then by Theorem $6.2 .6$ it is included in a hyperplane $H=f^{-1}{c}$ for some non-zero linear form $f$ and $c \in \mathbb{R}$. Then
$$H={x: f(x) \geq c} \cap{x: f(x) \leq c},$$
an intersection of two half-spaces. Also,
$$H=u+V:={u+v: v \in V}$$

Let $V$ be a real vector space and $C$ a convex set in $V$. A real-valued function $f$ on $C$ is called convex iff
$$f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)$$

## MATH20101COURSE NOTES ：

$$t \leq 3 \sum_{i=1}^{q} \lambda\left(J_{i}\right) \leq 6 j \sum_{i=1}^{q} \mu\left(J_{i}\right) \leq 6 j \mu(V) \leq 6 j \varepsilon .$$
Thus $\lambda\left(P_{j}\right) \leq 6 j \varepsilon$. Letting $\varepsilon \downarrow 0$ gives $\lambda\left(P_{j}\right)=0, j=1,2, \ldots$, and letting $j \rightarrow \infty$ proves the lemma.
Now to prove Theorem 7.2.1, for each rational $r$ let
$$g_{r}:=\max (g-r, 0), \quad f_{r}(x):=\int_{a}^{x} g_{r}(t) d t .$$

# 实分析| Real Analysis代写 MT3502

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and $c_{n} \rightarrow a, d_{n} \rightarrow b$; then
$$F\left(d_{n}\right)-F\left(c_{n}\right)=H\left(d_{n}\right)-H\left(c_{n}\right)$$
for all $n$, so in the limit we have
$$F(b)-0=H(b)-0$$

by the continuity of $F$ and In other words,
$$\int_{a}^{b} f=\int_{a}^{b} f \cdot \diamond$$

## MT3502 COURSE NOTES ：

Let $F:[a, b] \rightarrow \mathbb{R}$ and $G:[a, b] \rightarrow \mathbb{R}$ be the indefinite upper integrals of $f$ and $g$, respectively:
$$F(x)=\int_{a}^{-x} f \text { and } G(x)=\int_{a}^{-x} g$$
for $a \leq x \leq b$. If $a<c<d<b$ then, as in the proof of $9.6 .3$,
\begin{aligned} &F(d)=F(c)+\int_{c}^{-d} f, \ &G(d)=G(c)+\int^{-d} g \end{aligned}

# 实分析|Real Analysis  5CCM221A

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Proof. Let $h_{n} \rightarrow 0, h_{n} \neq 0$. Since $E$ is continuous, $E\left(h_{n}\right) \rightarrow E(0)=$ 1. Let $x_{n}=E\left(h_{n}\right)$; then $x_{n}>0, x_{n} \rightarrow 1$ and $x_{n} \neq 1$ (because $\left.h_{n} \neq 0\right)$, so by $9.5 .10$
$$\frac{L\left(x_{n}\right)}{x_{n}-1} \rightarrow 1$$

That is,
$$\frac{h_{n}}{E\left(h_{n}\right)-1} \rightarrow 1$$
whence the lemmn.

## 5CCM221ACOURSE NOTES ：

(i) Show that $A$ is an odd function: $A(-x)=-A(x)$ for all $x \in \mathbb{R}$.
{Hint: $f$ is an even function.}
(ii) $A$ is strictly increasing. {Hint: $\left.A^{\prime}=f .\right}$
(iii) For every positive integer $k$,
$$\frac{1}{1+k^{2}} \leq A(k)-A(k-1) \leq \frac{1}{1+(k-1)^{2}} .$$
{Hint: Integrate $f$ over the interval
(iv) Let
$$s_{n}=\sum_{k=1}^{n} \frac{1}{1+k^{2}} .$$

# 实分析|Real Analysis代写 MATH 3150

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Proof. Write $h=g \circ f .$ By 8.1.5, there exists a function $A: S \rightarrow \mathbb{R}$, continuous at $c$, such that
$$f(x)-f(c)=A(x)(x-c) \text { for all } x \in \mathbf{S}$$
Similarly, there is a function $B: T \rightarrow \mathbb{R}$, continuous at $f(c)$, such that

(2) $g(y)-g(f(c))=B(y)(y-f(c))$ for all $y \in \mathrm{T}$.
If $s \in \mathrm{S}$ then $f(x) \in \mathrm{T}$; putting $y=f(x)$ in (2), we have
\begin{aligned} g(f(x))-g(f(c)) &=B(f(x)) \cdot(f(x)-f(c)) \ &=B(f(x)) \cdot A(x)(x-c) \end{aligned}

$$\sigma={a<b}, \quad \tau \doteq{a<c<b}$$
Writing $M=\sup f$ as before, and
\begin{aligned} M^{\prime} &=\sup {f(x): a \leq x \leq c}, \ M^{\prime \prime} &=\sup {f(x): c \leq x \leq b}, \end{aligned}
$$S(\sigma)=M(b-a) \quad \text { and } \quad S(\tau)=M^{\prime}(c-a)+M^{\prime \prime}(b-c)$$