ANA1_L28

Using Sample points instead of M and m

Rudin, 6.7

Theorem 1(Theorem).

Let $f:[a,b]\to \mathbb{R}$. If $U(P,f,\alpha )-L(P,f,\alpha )<ϵ$ holds for some $P$, and if $s_i$ and $t_i$ are arbitrary points in $[x_{i-1},x_i]$, then

$$\sum _{i=1}^n|f(s_i)-f(t_i)|\Delta {\alpha }_i<ϵ.$$

If $f\in \mathcal{R}(\alpha )$, then

$$\left|\sum _{i=1}^{n}f(t_i)\Delta {\alpha }_i-{\int }_a^{b}fd\alpha \right|<ϵ.$$

Proof
The first part’s trivial. The inequalities

$$L(P,f,\alpha )\le \sum f(t_i)\Delta {\alpha }_i\le U(P,f,\alpha )$$

and

$$L(P,f,\alpha )\le {\int }_a^{b}fd\alpha \le U(P,f,\alpha )$$

prove the second part. ❏

---

Recall

Integrability of bounded $f$ on $[a,b]$:

$f$ continuous $⟹$ $f\in \mathcal{R}(\alpha )$.
$f$ monotonic and $\alpha$ continuous $⟹$ $f\in \mathcal{R}(\alpha )$.
$f$ has finitely many discontinuities at which $\alpha$ is continuous $⟹$ $f\in \mathcal{R}(\alpha )$.
Compositions of integrable functions with continuous functions are continuous.

---

Integration and differentiation

Rudin, 6.20

Theorem 2(Theorem).

Let $f\in \mathcal{R}$ on $[a,b]$. For $a\le x\le b$, define

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

Then $F$ is continuous on $[a,b]$. Furthermore, if $f$ is continuous at a point $x_0$ of $[a,b]$, then $F$ is differentiable at $x_0$, and

$$F^{\prime }(x_0)=f(x_0).$$

Proof
For $f\in \mathcal{R}$ to be true, $f$ must be bounded. Suppose $|f(t)|\le M$ in $[a,b]$. We will prove that $F$ is uniformly continuous (uniform continuity implies continuity). Let $a\le x<y\le b$. Let $ϵ>0$. Then, if $y-x<\frac{ϵ}{M}$, we have

$$|F(y)-F(x)|=\left|{\int }_x^{y}f(t)dt\right|<M(y-x)=ϵ.$$

Now, suppose $f$ is continuous as $x_0$. Let $ϵ>0$. Choose $\delta >0$ such that $|t-x_0|<\delta$ $⟹$ $|f(t)-f(x_0)|<ϵ$. Let $|t-x_0|<\delta$.

$$\begin{array}{rl}\left|\frac{1}{t-x_0}{\int }_{x_0}^{t}f(u)du-f(x_0)\right|& =\left|\frac{1}{t-x_0}\right|\left|{\int }_{x_0}^t[f(u)-f(x_0)]du\right|\\ & \le ϵ.\end{array}$$

Thus, $F^{\prime }(x_0)=f(x_0)$. ❏

Rudin drags his feet in the last part of the proof, since he wants to avoid integrals where the limits of integration are not in the right order, since technically those aren’t defined.

The fun theorem

Rudin, 6.21

Theorem 3(Theorem).

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 $ϵ>0$. Since $f\in \mathcal{R}$ on $[a,b]$, we can choose a partition $P$ of $[a,b]$ such that $U(P,f)-L(P,f)<ϵ$. Because of the mean value theorem, we can choose $t_i\in [x_{i-1},x_i]$ such that

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

We know from Rudin, 6.7 that

$$\left|\sum _{i=1}^{n}f(t_i)\Delta x_i-{\int }_a^{b}fdx\right|<ϵ$$

Thus,

$$\left|F(b)-F(a)-{\int }_a^{b}fdx\right|<ϵ.$$

❏

---

Integration by parts

We’ll need this theorem:

Product of integrable functions is integrable

Rudin, 6.13a

Theorem 4(Theorem).

If $f\in \mathcal{R}$ and $g\in \mathcal{R}$ on $[a,b]$, then $fg\in \mathcal{R}$.

Proof
We know that $f+g$ and $f-g$ are integrable on $[a,b]$, from 6.12a. We can compose these with $x^2$ to obtain integrable functions $(f+g{)}^2$ and $(f-g{)}^2$. It follows that $(f+g{)}^2+(f-g{)}^2=4fg$ is integrable. ❏

Rudin, 6.22

Theorem 5(Theorem).

Suppose $F$ and $G$ 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 $H(x)=F(x)G(x)$. Then, $h\equiv H^{\prime }=fG+Fg$. Note that since $f,g\in \mathcal{R}$ and $F,G\in \mathcal{R}$(since differentiability implies continuity and continuous functions are integrable), $h\in \mathcal{R}$. From the fundamental theorem, we have

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

❏

---

Misc

Rudin, 6.13b

Theorem 6(Theorem).

If $f\in \mathcal{R}(\alpha )$, then

$$|f|\in \mathcal{R}(\alpha )\text{ and }\left|{\int }_a^{b}fd\alpha \right|\le {\int }_a^b|f|d\alpha .$$

Proof
Composing $f$ with $|x|$ yields $|f|\in \mathcal{R}(\alpha )$.
Let $c=\pm 1$, such that $c\int fd\alpha \ge 0$. Then,

$$\left|\int fd\alpha \right|=c\int fd\alpha =\int cfd\alpha \le \int |f|d\alpha$$

since $cf\le |f|$. ❏