Fundamental Theorem of Calculus/First Part/Proof 1

Proof
To show that $F$ is a primitive of $f$ on $\closedint a b$, we need to establish the following:


 * $F$ is continuous on $\closedint a b$
 * $F$ is differentiable on the open interval $\openint a b$
 * $\forall x \in \closedint a b: \map {F'} x = \map f x$.

Proof that $F$ is Continuous
We have that $f$ is continuous on $\closedint a b$.

It follows from Continuous Image of Closed Interval is Closed Interval that $f$ is bounded on $\closedint a b$.

Suppose that:
 * $\forall t \in \closedint a b: \size {\map f t} < \kappa$

Let $x, \xi \in \closedint a b$.

From Sum of Integrals on Adjacent Intervals for Continuous Functions‎, we have that:
 * $\displaystyle \int_a^x \map f t \rd t + \int_x^\xi \map f t \rd t = \int_a^\xi \map f t \rd t$

That is:
 * $\displaystyle \map F x + \int_x^\xi \map f t \rd t = \map F \xi$

So:
 * $\displaystyle \map F x - \map F \xi = -\int_x^\xi \map f t \rd t = \int_\xi^x \map f t \rd t$

From the corollary to Upper and Lower Bounds of Integral:
 * $\size {\map F x - \map F \xi} < \kappa \size {x - \xi}$

Thus it follows that $F$ is continuous on $\closedint a b$.

Proof that $F$ is Differentiable and $f$ is its Derivative
It is now to be shown that that $F$ is differentiable on $\openint a b$ and that:
 * $\forall x \in \closedint a b: \map {F'} x = \map f x$

Let $x, \xi \in \closedint a b$ such that $x \ne \xi$.

Then:

Now, let $\epsilon > 0$.

If $\xi \in \openint a b$, then $f$ is continuous at $\xi$.

So for some $\delta > 0$:
 * $\size {\map f t - \map f \xi} < \epsilon$

provided $\size {t - \xi} < \delta$.

So provided $\size {x - \xi} < \delta$ it follows that:
 * $\size {\map f t - \map f \xi} < \epsilon$

for any $t$ in an interval whose endpoints are $x$ and $\xi$.

So from the corollary to Upper and Lower Bounds of Integral, we have:

provided $0 < \size {x - \xi} < \delta$.

But that is what this means:


 * $\dfrac {\map F x - \map F \xi} {x - \xi} \to \map f \xi$ as $x \to \xi$

So $F$ is differentiable on $\openint a b$, and:
 * $\forall x \in \closedint a b: \map {F'} x = \map f x$