265 lines
8.1 KiB
Markdown
265 lines
8.1 KiB
Markdown
# Math4121 Exam 1 Review
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Range: Chapter 5 and 6 of Rudin. We skipped (and so you will not be tested on)
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- Differentiation of Vector Valued Functions (pp. 111-113)
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- Integration of Vector-Valued Function and Rectifiable Curves (pp.135-137)
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You will also not be tested on Uniform Convergence and Integration, which we cover in class on Monday 2/10.
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## Chapter 5: Differentiation
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### Definition of the Derivative
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Let $f$ be a real function defined on an closed interval $[a,b]$. We say that $f$ is differentiable at a point $x \in [a,b]$ if the following limit exists:
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$$
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f'(x) = \lim_{t\to x} \frac{f(t) - f(x)}{t - x}
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$$
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If the limit exists, we call it the derivative of $f$ at $x$ and denote it by $f'(x)$.
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#### Theorem 5.2
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Every differentiable function is [continuous](https://notenextra.trance-0.com/Math4111/Math4111_L22#definition-45).
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The converse is not true, consider $f(x) = |x|$.
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#### Theorem 5.3
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If $f,g$ are differentiable at $x$, then
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1. $(f+g)'(x) = f'(x) + g'(x)$
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2. $(fg)'(x) = f'(x)g(x) + f(x)g'(x)$
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3. If $g(x) \neq 0$, then $(f/g)'(x) = (f'(x)g(x) - f(x)g'(x))/g(x)^2$
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#### Theorem 5.4
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Constant function is differentiable and its derivative is $0$.
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#### Theorem 5.5
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Chain rule: If $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$, then the composite function $g\circ f$ is differentiable at $x$ and
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$$
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(g\circ f)'(x) = g'(f(x))f'(x)
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$$
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#### Theorem 5.8
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The derivative of local extremum ($\exists \delta > 0$ s.t. $f(x)\geq f(y)$ or $f(x)\leq f(y)$ for all $y\in (x-\delta,x+\delta)$) is $0$.
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#### Theorem 5.9
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Generalized mean value theorem: If $f,g$ are differentiable on $(a,b)$, then there exists a point $x\in (a,b)$ such that
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$$
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(f(b)-f(a))g'(x) = (g(b)-g(a))f'(x)
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$$
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If we put $g(x) = x$, we get the mean value theorem.
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$$
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f(b)-f(a) = f'(x)(b-a)
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$$
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for some $x\in (a,b)$.
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#### Theorem 5.12
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Intermediate value theorem:
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If $f$ is differentiable on $[a,b]$, for all $\lambda$ between $f'(a)$ and $f'(b)$, there exists a $c\in (a,b)$ such that $f'(x) = \lambda$.
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#### Theorem 5.13
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L'Hôpital's rule: If $f,g$ are differentiable in $(a,b)$ and $g'(x) \neq 0$ for all $x\in (a,b)$, where $-\infty \leq a < b \leq \infty$,
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Suppose
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$$
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\frac{f'(x)}{g'(x)} \to A \text{ as } x\to a
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$$
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If
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$$
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f(x) \to 0, g(x) \to 0 \text{ as } x\to a
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$$
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or if
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$$
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g(x) \to \infty \text{ as } x\to a
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$$
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then
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$$
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\lim_{x\to a} \frac{f(x)}{g(x)} = A
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$$
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#### Theorem 5.15
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Taylor's theorem: If $f$ is $n$ times differentiable on $[a,b]$, $f^{(n-1)}$ is continuous on $[a,b]$, and $f^{(n)}$ exists on $(a,b)$, for any distinct points $\alpha, \beta \in [a,b]$, there exists a point $x\in (\alpha, \beta)$ such that
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$$
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f(\beta) =\left(\sum_{k=0}^{n-1} \frac{f^{(k)}(\alpha)}{k!}(\beta-\alpha)^k\right) + \frac{f^{(n)}(x)}{n!}(\beta-\alpha)^n
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$$
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## Chapter 6: Riemann-Stieltjes Integration
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### Definition of the Integral
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Let $\alpha$ be a monotonically increasing function on $[a,b]$.
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A partition of $[a,b]$ is a set of points $P = \{x_0, x_1, \cdots, x_n\}$ such that
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$$
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a = x_0 < x_1 < \cdots < x_n = b
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$$
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Let $\Delta \alpha_i = \alpha(x_{i}) - \alpha(x_{i-1})$ for $i = 1, \cdots, n$.
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Let $m_i = \inf \{f(x) : x_{i-1} \leq x \leq x_{i}\}$ and $M_i = \sup \{f(x) : x_{i-1} \leq x \leq x_{i}\}$ for $i = 1, \cdots, n$.
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The lower sum of $f$ with respect to $\alpha$ is
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$$L(f,P,\alpha) = \sum_{i=1}^{n} m_i \Delta \alpha_i$$
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The upper sum of $f$ with respect to $\alpha$ is
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$$U(f,P,\alpha) = \sum_{i=1}^{n} M_i \Delta \alpha_i$$
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Let $\overline{\int_a^b} f(x) d\alpha(x)=\sup_P L(f,P,\alpha)$ and $\underline{\int_a^b} f(x) d\alpha(x)=\inf_P U(f,P,\alpha)$.
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If $\overline{\int_a^b} f(x) d\alpha(x) = \underline{\int_a^b} f(x) d\alpha(x)$, we say that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ and we write
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$$
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\int_a^b f(x) d\alpha(x) = \overline{\int_a^b} f(x) d\alpha(x) = \underline{\int_a^b} f(x) d\alpha(x)
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$$
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#### Theorem 6.4
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Refinement of partition will never make the lower sum smaller or the upper sum larger.
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$$
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L(f,P,\alpha) \leq L(f,P^*,\alpha) \leq U(f,P^*,\alpha) \leq U(f,P,\alpha)
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$$
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#### Theorem 6.5
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$\underline{\int_a^b} f(x) d\alpha(x) \leq \overline{\int_a^b} f(x) d\alpha(x)$
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#### Theorem 6.6
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$f\in \mathscr{R}(\alpha)$ on $[a,b]$ if and only if for every $\epsilon > 0$, there exists a partition $P$ of $[a,b]$ such that
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$$
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U(f,P,\alpha) - L(f,P,\alpha) < \epsilon
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$$
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#### Theorem 6.8
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Every continuous function on a closed interval is Riemann-Stieltjes integrable with respect to any monotonically increasing function.
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#### Theorem 6.9
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If $f$ is monotonically increasing on $[a,b]$ and **$\alpha$ is continuous on $[a,b]$**, then $f\in \mathscr{R}(\alpha)$ on $[a,b]$.
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Key: We can repartition the interval $[a,b]$ using $f$.
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#### Theorem 6.10
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If $f$ is bounded on $[a,b]$ and has only **finitely many discontinuities** on $[a,b]$, then $f\in \mathscr{R}(\alpha)$ on $[a,b]$.
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Key: We can use the bound and partition around the points of discontinuity to make the error arbitrary small.
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#### Theorem 6.11
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If $f\in \mathscr{R}(\alpha)$ on $[a,b]$, and $m\leq f(x) \leq M$ for all $x\in [a,b]$, and $\phi$ is a continuous function on $[m,M]$, then $\phi\circ f\in \mathscr{R}(\alpha)$ on $[a,b]$.
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_Composition of bounded integrable functions and continuous functions is integrable._
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#### Theorem 6.12
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Properties of the integral:
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Let $f,g\in \mathscr{R}(\alpha)$ on $[a,b]$, and $c$ be a constant. Then
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1. $f+g\in \mathscr{R}(\alpha)$ on $[a,b]$ and $\int_a^b (f(x) + g(x)) d\alpha(x) = \int_a^b f(x) d\alpha(x) + \int_a^b g(x) d\alpha(x)$
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2. $cf\in \mathscr{R}(\alpha)$ on $[a,b]$ and $\int_a^b cf(x) d\alpha(x) = c\int_a^b f(x) d\alpha(x)$
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3. $f\in \mathscr{R}(\alpha)$ on $[a,b]$ and $c\in [a,b]$, then $\int_a^b f(x) d\alpha(x) = \int_a^c f(x) d\alpha(x) + \int_c^b f(x) d\alpha(x)$.
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4. **Favorite Estimate**: If $|f(x)| \leq M$ for all $x\in [a,b]$, then $\left|\int_a^b f(x) d\alpha(x)\right| \leq M(\alpha(b)-\alpha(a))$.
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5. If $f\in \mathscr{R}(\beta)$ on $[a,b]$, then $\int_a^b f(x) d(\alpha+\beta) = \int_a^b f(x) d\alpha + \int_a^b f(x) d\beta$.
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#### Theorem 6.13
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If $f,g\in \mathscr{R}(\alpha)$ on $[a,b]$, then
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1. $fg\in \mathscr{R}(\alpha)$ on $[a,b]$
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2. $|f|\in \mathscr{R}(\alpha)$ on $[a,b]$ and $\left|\int_a^b f(x) d\alpha(x)\right| \leq \int_a^b |f(x)| d\alpha(x)$
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Key: (1), use Theorem 6.12, 6.11 to build up $fg$ from $(f+g)^2-f^2-g^2$. (2), take $\phi(x) = |x|$ in Theorem 6.11.
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#### Theorem 6.14
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Integration over indicator functions:
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If $a<s<b$, $f$ is bounded on $[a,b]$, and $f$ is continuous at $s$, and $\alpha(x)=I(x-s)$, then
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$$
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\int_a^b f(x) d\alpha(x) = f(s)
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$$
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Key: Note the max difference can be made only occurs at $s$.
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#### Theorem 6.15
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Integration over step functions:
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If $\alpha(x) = \sum_{i=1}^{n} c_i I(x-x_i)$ for $x\in [a,b]$, then
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$$
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\int_a^b f(x) d\alpha(x) = \sum_{i=1}^{n} c_i f(x_i)
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$$
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#### Theorem 6.21
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Fundamental theorem of calculus:
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Let $f\in \mathscr{R}(\alpha)$ on $[a,b]$, and $F(x) = \int_a^x f(t) d\alpha(t)$. Then
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1. $F$ is continuous on $[a,b]$
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2. If $f$ is continuous at $x\in [a,b]$, then $F$ is differentiable at $x$ and $F'(x) = f(x)$
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## Chapter 7: Sequence and Series of Functions
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### Example of non-Riemann integrable function
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$\lim_{m\to \infty} \lim_{n\to \infty} (\cos(m!\pi x))^{2n}=\begin{cases} 1 & x\in \mathbb{Q} \\ 0 & x\notin \mathbb{Q} \end{cases}$
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This function is everywhere discontinuous and not Riemann integrable.
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### Uniform Convergence
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#### Definition 7.7
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A sequence of functions $\{f_n\}$ converges uniformly to $f$ on $E$ if for every $\epsilon > 0$, there exists a positive integer $N$ such that
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$$
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|f_n(x) - f(x)| < \epsilon \text{ for all } x\in E \text{ and } n\geq N
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$$
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If $E$ is a point, then that's the common definition of convergence.
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If we have uniform convergence, then we can swap the order of limits.
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#### Theorem 7.16
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If $\{f_n\}\in \mathscr{R}(\alpha)$ on $[a,b]$, and $\{f_n\}$ converges uniformly to $f$ on $[a,b]$, then
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$$
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\int_a^b f(x) d\alpha(x) = \lim_{n\to \infty} \int_a^b f_n(x) d\alpha(x)
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$$
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Key: Use the definition of uniform convergence to bound the difference between the integral of the limit and the limit of the integral. $\int_a^b (f-f_n)d\alpha \leq |f-f_n| \int_a^b d\alpha = |f-f_n| (\alpha(b)-\alpha(a))$.
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