The Riemann integral functional is represented by Lebesgue measure on an interval

Linearity of the Riemann integral makes $L$ linear, and $f\ge 0$ implies $L(f)\ge 0$. Since $[a,b]$ is compact, $C_c([a,b])=C([a,b])$.

Under the stated choice hypothesis, Lebesgue measure is regular on [given] $\mathbb{R}$; its restriction to the closed subspace $[a,b]$ is finite and Radon. For every $f\in C([a,b])$, equality of the Riemann and Lebesgue integrals gives $L(f)={\int }_{[a,b]}fd(\lambda {\mathrm{\mid }}_{[a,b]})$. Thus $\lambda {\mathrm{\mid }}_{[a,b]}$ is a Radon representing measure.

To prove uniqueness without a general LCH cutoff, let $\rho$ be any other Radon representing measure on $[a,b]$. If $K\subseteq O\subseteq [a,b]$ with $K$ compact and $O$ relatively open, A compact set and a disjoint closed set have a positive norm-distance gap gives a positive distance between nonempty $K$ and the closed set $[a,b]\setminus O$; when the latter is empty use the constant function $1$, and when $K$ is empty the comparison is trivial. Choose a smaller $\delta >0$. The function $u(x)=\max \{0,1-d(x,K)\mathrm{/}\delta \}$ is continuous by $\mathrm{\mid }d(x,A)-d(y,A)\mathrm{\mid }\le d(x,y)$, so the distance to a fixed nonempty set is $1$-Lipschitz, equals $1$ on $K$, and vanishes outside $O$. Equality of the two representing integrals therefore gives $\lambda (K)\le \int ud\lambda =\int ud\rho \le \rho (O)$; symmetry gives $\rho (K)\le \lambda (O)$. Outer regularity yields equality on compact sets, inner regularity then equality on opens, and outer regularity yields equality on every Borel set. Thus $\rho =\lambda {\mathrm{\mid }}_{[a,b]}$.