(Theorem) Fundamental theorem of calculus

Theorem (Rudin 6.20): FTC I for a single variable

Suppose $f:[a,b]\to \mathbb{R}$ is Darboux integrable. Define $F:[a,b]\to \mathbb{R}$ by

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

Then $F$ is uniformly continuous on $[a,b]$. Moreover, if $f$ is continuous at $x_0\in [a,b]$, then $F$ is differentiable at $x_0$ with $F’(x_0)=f(x_0)$.

Proof from Modern Analysis I. Uses:

• (Proof pattern) Middle-man trick

First, we show $F$ is uniformly continuous. Since $f$ is Darboux integrable, it is also bounded on $[a,b]$, and there exists $M>0$ such that $|f(t)|\le M$ for all $t\in [a,b]$. For all $ϵ>0$, set $\delta =ϵ/M$. Then for all $a\le x\le y\le b$, we have

$$\begin{array}{rl}|x-y|<\delta ⟹& |F(y)-F(x)|=\left|{\int }_x^{y}f(t)dt\right|\\ & \le {\int }_x^y|f(t)|dt\le M|y-x|\le M\delta =ϵ.\end{array}$$

Now suppose $f$ is continuous at $x_0\in [a,b]$, and let $ϵ>0$. Then there exists some $\delta >0$ such that

$$|f(t)-f(x_0)|<ϵ$$

for all $t\in B(x_0,\delta )\cap [a,b]$. In particular, for all $t>x_0$, we have

$$\begin{array}{rl}\frac{F(t)-F(x_0)}{t-x_0}& =\frac{1}{t-x_0}{\int }_{x_0}^{t}f(x)dx=\frac{1}{t-x_0}{\int }_{x_0}^t|f(x)-f(x_0)|dx+f(x_0).\end{array}$$

By continuity of $f$ at $x_0$, we know

$$\frac{1}{t-x_0}{\int }_{x_0}^t|f(x)-f(x_0)|dx\le \frac{1}{t-x_0}{\int }_{x_0}^t|f(x)-f(x_0)|dx<ϵ,$$

so

$$\left|\frac{F(t)-F(x_0)}{t-x_0}-f(x_0)\right|<ϵ.$$

Since $ϵ>0$ is arbitrary, it follows from taking the limit as $t\to x_0$ that $F’(x_0)=f(x_0)$ as claimed. $\square$

Theorem (Rudin 6.20): FTC II for a single variable

Let $f:[a,b]\to \mathbb{R}$ be Darboux integrable, and suppose there exists a differentiable function $F:[a,b]\to \mathbb{R}$ such that $F’=f$. Then

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

Proof from Modern Analysis I.

Let $ϵ>0$. Since $f$ is Darboux integrable, there exists a partition $P=\{x_0,\dots ,x_n\}$ of $[a,b]$ such that the upper and lower sums satisfy

$$U(f,P)-L(f,P)<ϵ.$$

By the mean value theorem, for each $j\in 1\dots n$, there exists ${ϵ}_j\in [x_{j-1},x_j]$ such that

$$F(x_j)-F(x_{j-1})=f({ϵ}_j)(x_j-x_{j-1}).$$

Then we have

$$L_f\le F(b)-F(a)=\sum _{j=1}^{n}f({ϵ}_j)(x_j-x_{j-1})\le U_f,$$

and

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

as well. The equality follows since $ϵ>0$ is arbitrary. $\square$