Daniell integral vs. Borel measure Let $X$ be a locally compact, Hausdorff, topological space and denote by $\mathcal B_+(X)$ the collection of all Borel-measurable functions from $X$ to $[0,+\infty]$ (extended positive reals).
Suppose that we are given a map
  $$
  I:\mathcal B_+(X) \to [0,+\infty]
  $$
  which is positively homogeneous and satisfies
  $$
  I\big(\sum_{n=1}^\infty f_n\big) =   \sum_{n=1}^\infty I(f_n),
  $$
  for every sequence $\{f_n\}_n$ in $\mathcal B_+(X)$.
Observe that this is essentially the Daniell integral.
Suppose further that $\mu$ is a given regular Borel measure on $X$ such that
  $$
  I(f) = \int_X f\,d\mu,
  $$
  for every non-negative $f$ in $C_c(X)$, the space of all compactly supported continuous functions on $X$.  Does it follow that this identity also holds for every $f$ in $\mathcal B_+(X)$?
Ok to assume that $X$ is metrizable and $\sigma$-compact, if necessary.
 A: Here is an answer for $X$ metrizable and $\sigma$-compact. I assume $I$ is finite on compactly supported continuous functions. 
Recall that a locally compact metrizable $\sigma$-compact space is Polish (separable and completely metrizable). Moreover, there is an increasing sequence of compact sets $\langle K_n\rangle$ such that $K_n$ is a subset of the interior of $K_{n+1}$ for all $n$ and such that $X=\bigcup_n K_n$. Fix some metric $d$ that completely metrizes $X$.
Clearly, it suffices to show $I(1_A)=\mu(A)$ for every measurable set $A$. By the regularity of $\mu$, one only has to prove this for $A$ compact and, even nicer, for sets of the form $A\cap K_n$ with $A$ compact. Now there is going to be some $\epsilon$ such that $(A\cap K_n)_\epsilon=\{y\in X\mid d(y, A\cap K_n)<\epsilon\}$ is a subset of the interior of $K_{n+1}$. Define $f_n:X\to\mathbb{R}$ by $f_m(x)=\max\{1-md(x,A\cap K_n),0\}$. This is a continuous function and for $m$ larger than $1/\epsilon$, it vanishes outside the compact set $K_{n+1}$. Moreover $\langle f_m\rangle$ decreases pointwise to $1_{A\cap K_n}$. Since $I$ satisfies the dominated convergence theorem, the conclusion follows.
