Trouble understanding proof of a variant of the Riesz representation theorem I am trying to understand a proof of a variant of the Riesz representation theorem. 
Consider a linear functional $l$ on the space of continuous functions on $[0,1]$. Assume further that $\ell(f)\ge 0$ when $f\ge0$ on $[0,1]$.
We are going to define a function $F(u)$ on $[0,1]$ that will be used to construct a Lebesgue-Stieltjes measure. Define a collection of auxiliary functions $f_\epsilon(x,u)$ so they are $1$ on $[0,u]$, $0$ on $[u+\epsilon, 1]$, and linearly interpolated between those two intervals. The graph from left to right of such a function is a horizontal line with $y=1$, a downward sloping line, and then a horizontal line along 0.
Define $$F(u)=\lim_{\epsilon\rightarrow 0}\ \ell(f_\epsilon(x,u)).$$
We see that $F$ is increasing, and as shown in the answers, right-continuous. Why do we have
$$\ell(f)=\int_0^1 f(x)\ dF(x)$$ 
where the above is a Lebesgue integral.
 A: Right continuity of $F$ can be derived as follows. If $C = l(\mathbf{1})$ then $l(f) \leq C \cdot \max(f)$ for all continuous $f$ on $[0,1]$.  In particular if $f_n$ is a sequence of non-negative functions such that $\max(f_n) \to 0$ then $l(f_n) \to 0$. Use this on $f_{\varepsilon}(x,u + \varepsilon') - f_{\varepsilon}(x, u)$ to conclude that
$$\lim_{\varepsilon' \downarrow 0}l(f_{\varepsilon}(x, u + \varepsilon')) = l(f_{\varepsilon}(x,u)).
$$
Then for all $\delta > 0$ there exist $\varepsilon, \varepsilon' > 0$ such that
$$F(u) \leq F(u + \varepsilon') \leq l(f_{\varepsilon}(x, u + \varepsilon')) \leq l(f_{\varepsilon}(x, u)) + \frac{\delta}{2} \leq F(u) + \delta.
$$
A: If you look at the Riemann sums for $\int_0^1 f(x)dF(x)$, they are like
$$\tag{1}
\sum_j f(t_{j+1})\,(F(t_{j+1})-F(t_j))=\; \lim_{\varepsilon\to0}\; l\left(\sum_jf(t_{j+1})(f_\varepsilon(x,t_{j+1})-f_\varepsilon(x,t_j))\right)
$$
If you look carefully at the functions $x\mapsto \sum_jf(t_{j+1})(f_\varepsilon(x,t_{j+1})-f_\varepsilon(x,t_j))$, you'll notice they are piecewise linear functions that take the values $f(t_j)$ at $t_j$. So it is not hard to show that if the partitions for your Riemann sums are good (say, equally spaced) then these functions converge uniformly to $f$ (remember that $f$ is uniformly continuous).
