LEC ANA1 28

We can sample the value of the function in an interval instead of using the supremum/infimum:

Theorem 150.1(Rudin 6.7).

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.□

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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.

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Integration and differentiation

Theorem 150.2(Rudin 6.20).

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

Theorem 150.3(Rudin 6.21, 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$ϵ>0$. 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,

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

88e00c

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Integration by parts

Theorem 150.4(Rudin 6.22).

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

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

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Misc

Theorem 150.5(Rudin 6.13b).

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 $cf\le |f|$.□