Fundamental Theorem of Calculus

Proposition (Change of Variable)

First let $f$ be a bounded function from $[a,b]\to \mathbb{R}$ and let $\alpha :[a,b]\to \mathbb{R}$ be an increasing function. Let $\varphi :[c,d]\to [a,b]$ be a strictly increasing continuous function with $\varphi (c)=a$ and $\varphi (d)=b$.

Then $f\in \mathcal{R}(\alpha )⟺f\circ \varphi \in \mathcal{R}(\alpha \circ \varphi )$ and

$${\int }_a^{b}fd\alpha ={\int }_c^d(f\circ \varphi )d(\alpha \circ \varphi )$$

Note that $\varphi$ is a continuous bijection between $[a,b]$ and $[\alpha ,\beta ]$. Or, a homeomorphism. It is like a dictionary to connect two different domains. The composition $\alpha \circ \varphi$ gives us a new measure for any subinterval of the new domain.

Proof: We still show for any partition $P$ the set of Riemann-Stieltjes Sums for $f,\alpha$ are in one-to-one correspondence with the the sums for $f\circ \varphi ,\alpha \circ \varphi$ with respect to a partition of $[c,d]$. So, let

$$P:=\{x_1,x_2,\dots ,x_n\}$$

where

$$\sum _{i=1}^{n}f({\xi }_i)(\alpha (x_i)-\alpha (x_{i-1})){\xi }_i\in [x_{i-1},x_i]$$

a Riemann sum for $f$ and $\alpha$. Since $\varphi$ is a homeomorphism, $P$ corresponds with

$$P^{\prime }:=\{{\varphi }^{-1}(x_1),{\varphi }^{-1}(x_2),\dots ,{\varphi }^{-1}(x_n)\}$$

$\varphi$ needs to be increasing so $\varphi$ is a bijection.

in which each point is unique. Then, every ${\xi }_i$ has a correspondence to ${\varphi }^{-1}({\xi }_i)$ and we get the sum:

$$\sum _{i=1}^n(f\circ \varphi )({\varphi }^{-1}({\xi }_i))((\alpha \circ \varphi )({\varphi }^{-1}(x_i))-(\alpha \circ \varphi )({\varphi }^{-1}(x_{i-1})))$$

But then this form can be recognized as Riemann-Stieltjes Sum for $f\circ \varphi$ and $\alpha \circ \varphi$ on $P^{\prime }$. where

$$\begin{array}{rl}U(P,f,\alpha )& =U(P^{\prime },f\circ \varphi ,\alpha \circ \varphi )\\ L(P,f,\alpha )& =L(P^{\prime },f\circ \varphi ,\alpha \circ \varphi )\end{array}$$

and so the respective upper and lower integrals are equal.

If $\varphi$ is strictly decreasing, then the bounds order needs to change. However, the theorem still holds.

Theorem (First Fundamental Theorem of Calculus)

If $f:[a,b]\to \mathbb{R}$ is Riemann Integrable, define

$$F(x)={\int }_a^{x}f(t)dt$$

for $x\in [a,b]$. Then $F$ is continuous. Furthermore, if $f$ is continuous at $x_0$, then the derivative $F^{\prime }(x_0)$ exists. In particular, $F^{\prime }(x_0)=f(x_0)$.

Proof: Take $p\in [a,b]$. WTS that $F$ is continuous at $p$. Assume $|f|\le M$. Then if $|x-p|\le \epsilon /M$, where $a<x\le p$, we have

$$|F(x)-F(p)|=\left|{\int }_a^{x}f(dt)-{\int }_a^{p}fdt\right|=\left|{\int }_x^{p}fdt\right|\le M\cdot |p-x|<\epsilon /M\cdot M=\epsilon$$

by properties of the integral and so $F$ is continuous.

Assume $f$ is continuous at $x_0$. WTS that $F^{\prime }(x_0)=f(x_0)$ where

$$\underset{x\to x_0}{\lim }\left|\frac{F(x)-F(x_0)}{x-x_0}-f(x_0)\right|=0$$

The absolute value is not actually needed here. (Derivative)

Then the sum is

$$\begin{array}{rl}\underset{x\to x_0}{\lim }\left|\frac{{\int }_{x_0}^{x}f(t)dt}{x-x_0}-\frac{{\int }_{x_0}^{x}f(x_0)dt}{x-x_0}\right|& =\underset{x\to x_0}{\lim }\left|\frac{{\int }_{x_0}^x(f(t)-f(x))dt}{x-x_0}\right|\\ & \le \underset{x\to x_0}{\lim }\underset{t\in [x_0,x]}{\sup }|f(t)-f(x_0)|=0\end{array}$$

which converges to $0$ when $x\to x_0,t\to x_0$.

Theorem (Second Fundamental Theorem of Calculus)

If $f\in \mathcal{R}$ on $[a,b]$ and if there is a differentiable function $F$ on $[a,b]$ such that $F^{\prime }=f$, then

$${\int }_a^{b}f(x)dx=F(b)-F(a)$$

Proof: Let $\epsilon >0$ be given. Choose a partition $P=\{x_0,\dots ,x_n\}$ of $[a,b]$ so that $U(P,f)-L(P,f)<\epsilon$. The MVT gives points $t_i\in [x_{i-1},x_i]$ such that

$$F(x_i)-F(x_{i-1})=f(t_i)\Delta x_i$$

for $i=1,\dots ,n$. Thus

$$\sum _{i=1}^{n}f(t_i)\Delta x_i=F(b)-F(a)$$

It follows from Rudin Theorem 6.7 that

$$\left|F(b)-F(a)-{\int }_a^{b}f(x)dx\right|<\epsilon$$

Since this is true for every $\epsilon >0$, we are done.

Need to fill in Theorem 6.7.

Theorem (Integration by Parts)

Suppose $F,G$ are differentiable functions on $[a,b]$, $F^{\prime }=f\in \mathcal{R}$, and $G^{\prime }=g\in \mathcal{R}$. Then

$${\int }_a^{b}F(x)g(x)dx=F(b)G(b)-F(a)G(a)-{\int }_a^{b}f(x)G(x)dx$$

Proof: Let $H(x):=F(x)G(x)$. By Theorem (Second Fundamental Theorem of Calculus) to $H$ and its derivative, we note that $H^{\prime }\in \mathcal{R}$ by Properties of the Integral.

$H^{\prime }(x)=F^{\prime }(x)G(x)+F(x)G^{\prime }(x)=fG+Fg$.