## 515 – The fundamental theorem of calculus

Suppose that $f\in{\mathcal R}[a,b]$ and $f$ has an antiderivative $G$. Then

$\displaystyle G(b)-G(a)=\int_a^b f(t)dt.$

Note we are assuming $f$ is Riemann integrable. This means that given $\epsilon>0$ we can find an $\eta>0$ such that if $P$ is a tagged partition of ${}[a,b]$ and $\Delta(P)<\eta$, then

$\displaystyle \left|R(f,P)-\int_a^b f(t)dt\right|<\epsilon.$

Recall that a tagged partition $P$ consists of a partition $\Phi$ of ${}[a,b]$, represented by a finite sequence of points

$a=x_0,

and a sequence of representatives of the intervals defined by this partition, i.e., a collection of points $x_i^*\in[x_{i-1},x_i]$ for $i=1,\dots,n.$

We denote by $\delta x_i$ the number $x_i-x_{i-1}$ and by $\Delta(P)=\Delta(\Phi)$ the norm of $P$,

$\displaystyle \Delta(P)=\max_{i=1,\dots,n}\delta x_i.$

The Riemann sum associated to $f$ and $P$ is the sum

$\displaystyle R(f,P)=\sum_{i=1}^n f(x_i^*)\delta x_i.$

Pick any partition $\Phi=\{a=x_0 of ${}[a,b]$ with $\Delta(\Phi)<\eta$. We want to define a particular tagged partition $P$ with underlying partition $\Phi$ by appealing to the fact that $f$ has an antiderivative $G$. Specifically, by the mean value theorem, we have that for all $i=1,\dots,n$, there is some $x_i^*\in[x_{i-1},x_i]$ such that

$G(x_i)-G(x_{i-1})=f(x_i^*)\delta x_i.$

Define $P$ in terms of the points $x_i^*$ and the partition $\Phi$. Then

$R(f,P)=\sum_{i=1}^n f(x_i^*)\delta x_i=\sum_{i=1}^n (G(x_i)-G(x_{i-1}))$ $\displaystyle =G(b)-G(a).$

By our choice of $\eta$, we know that $|R(f,P)-\int_a^b f(t)dt|<\epsilon$, so

$\displaystyle \left|G(b)-G(a)-\int_a^b f(t)dt\right|<\epsilon.$

But the left hand side is independent of $\epsilon$, and $\epsilon$ was arbitrary. It follows that $\int_a^b f(t)dt=G(b)-G(a)$, as we wanted.

The same argument, but restricting ourselves to the interval ${}[a,x]$, shows that

$\displaystyle G(x)-G(a)=\int_a^x f(t)dt = F(x) -F(a)$,

where $F(x)=\int_a^x f(t)dt$ for $x\in[a,b]$, so in particular $F(a)=0$. It follows that $G$ and $F$ differ by a constant, and therefore, if $f$ is Riemann integrable and has an antiderivative at all, then $F$ is such an antiderivative.

The question remains of what Riemann integrable functions $f$ (with the intermediate value property) have antiderivatives. Another natural question has to do with the fact that our definition of antiderivative is very restrictive; it also makes sense to simply ask whether the equality $F'(x)=f(x)$ must hold for some $x$, assuming only that $f$ is integrable. It turns out that both questions require the introduction of the Lebesgue integral to be answered in a satisfactory way.