Proof: We use the Weierstrass approximation theorem (click for an alternative ad hoc proof). According to [6.7.2] there is a sequence of polynomials $({p}_{n})$ that converges uniformely on $[0,1]$ to f:
${p}_{n}\underset{gm}{\to}f$
As every polynomial $p}_{n$ is integrable due to [8.1.10], the uniform limit f has a primitive according to [8.1.15].
Proof: For each $n\in {\mathbb{N}}^{*}$ we will construct a traverse $p}_{n$ (a much shorter proof directly uses the Bernstein polynomials) that connects $n+1$ suitable points of f. To that end we section the interval $[\mathrm{0,1}]$ into n subintervals of length $\frac{1}{n}$ and define for $0\le i\le n1$
$I}_{n,i$
These subintervals correspond to $n+1$ points of f. We now connect the ith one to the ($i+1$)th one by the line segment
${g}_{n,i}:{I}_{n,i}\to \mathbb{R},\text{\hspace{1em}}x\mapsto {m}_{n,i}\cdot (x\frac{i}{n})+f(\frac{i}{n})$
where the slope $m}_{n,i$ is calculated by the quotient
${m}_{n,i}=\frac{f(\frac{i+1}{n})f(\frac{i}{n})}{\frac{1}{n}}=n(f(\frac{i+1}{n})f(\frac{i}{n}))$.
For $0\le i<n1$ we have $I}_{n,i}\cap {I}_{n,i+1}=\{\frac{i+1}{n}\$ and
${g}_{n,i}(\frac{i+1}{n})={m}_{n,i}\cdot \frac{1}{n}+f(\frac{i}{n})=f(\frac{i+1}{n})f(\frac{i}{n})+f(\frac{i}{n})={g}_{n,i+1}(\frac{i+1}{n})$,
so that we can take the well defined
concatenation
i 
For $f:A\to \mathbb{R}$ and $g:B\to \mathbb{R}$ with $f(x)=g(x)$ for all $x\in A\cap B$ we set
$f\cup g(x)=\{\begin{array}{l}f(x)\text{, if}x\in A\hfill \\ g(x)\text{, if}x\in B\hfill \end{array}$

$p}_{n}\u2254{g}_{n\mathrm{,0}}\cup \dots \cup {g}_{n,n1$
as the traverse we were looking for. A callable
applet
i 

exemplifies by the function $f:[0,1]\to \mathbb{R}$, $x\mapsto 5x\cdot \mathrm{cos}(20x)$ how the sequence of traverses approximates the function f.
Every line segment $g}_{n,i$ is the restriction of a linear function and hence integrable (see [8.1.10], [8.1.13]). Along with the line segments the traverses $p}_{n$ are integrable as well due to [8.1.14]. The desired integrability of f thus will follow from [8.1.15] if we could prove the uniform convergence
${p}_{n}\underset{uf}{\to}f$.
To that end let $x\in [\mathrm{0,1}]$ and $\epsilon >0$ be arbitrary. As f is uniformely continuous on the closed interval $[0,1]$ there is an $n}_{0}\in {\mathbb{N}}^{\ast$ such that all $r,s\in [0,1]$ satisfy the condition
$rs\le \frac{1}{{n}_{0}}\text{\hspace{1em}}\Rightarrow \text{\hspace{1em}}f(r)f(s)<\frac{\epsilon}{2}$.
For any $n\ge {n}_{0}$ we use this implication twice:

As $\frac{i+1}{n}\frac{i}{n}=\frac{1}{n}\le \frac{1}{{n}_{0}}$ the slopes of the line segments could be estimated as follows:
${m}_{n,i}=nf(\frac{i+1}{n})f(\frac{i}{n})<n\frac{\epsilon}{2}$ for all $0\le i<n1$.

x is a member of one of the intervals $I}_{n,i$, say $x\in {I}_{n,{i}_{n}}$. Thus we have ${p}_{n}(x)={g}_{n,{i}_{n}}(x)$ and $\frac{{i}_{n}}{n}x\le \frac{1}{n}\le \frac{1}{{n}_{0}}$, so we know:
$\begin{array}{ll}{p}_{n}(x)f(x)\hfill & ={g}_{n,{i}_{n}}(x)f(x)\hfill \\ \hfill & ={m}_{n,{i}_{n}}\cdot (x\frac{{i}_{n}}{n})+f(\frac{{i}_{n}}{n})f(x)\hfill \\ \hfill & \le {m}_{n,{i}_{n}}\cdot x\frac{{i}_{n}}{n}+f(\frac{{i}_{n}}{n})f(x)\hfill \\ \hfill & <n\cdot \frac{\epsilon}{2}\cdot \frac{1}{n}+\frac{\epsilon}{2}=\epsilon \text{.}\hfill \end{array}$