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.



Daniell integral vs. Borel measure
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.


