Representing sums of linear functionals on locally compact Hausdorff space using Riesz representation theorem. Suppose $\Lambda_1, \Lambda_2$ are positive linear functionals on $C_c(X)$, where $X$ a locally compact Hausdorff space $X$ then by Riesz Representation theorem for each one of them there's a sigma algebra ($\mathcal{M}_1$ and $\mathcal{M}_2$) and two regular measures ($\mu_1,\mu_2$ say) such that
$$\Lambda_i f = \int_X f d\mu_i \;\; i = 1, 2$$
Now $\Lambda = \Lambda_1 + \Lambda_2$ is also positive and linear so there's a sigma algebra $\mathcal{M}$ and a regular measure $\mu$ such that
$$
\left(\Lambda_1 + \Lambda_2 \right) f = \int_X f d\mu
$$
I am sort of guessing that $\mu$ might be $\mu = \mu_1 + \mu_2$ and $\mathcal{M} = \mathcal{M}_1 \cup \mathcal{M}_2$ but I am not entirely sure because of the two sigma algebras and the regularity properties would hold in this case.
Can anyone tell me if my guess is right or wrong?
 A: Regarding your guess: $\mathcal M_1\cup \mathcal M_2$ is not in general a $\sigma$-algebra (unless one contains the other), and even if it is, $\mu_1+\mu_2$ is not defined on all of $\mathcal M_1\cup \mathcal M_2$.
What you can do instead is take $\mathcal M=\mathcal M_1\cap \mathcal M_2$ and $\mu_1|_{\mathcal M}+\mu_2|_{\mathcal M}$. Then you can check that everything works.
However, there is (in my opinion) a better way to see this. Notice that the $\sigma$-algebras you have always contain the Borel $\sigma$-algebra (this is a part of the construction in Rudin's book; besides, for sufficiently simple ($\sigma$-compact) $X$ this is actually necessary in order to integrate all compactly supported continuous functions).
On the other hand, if $\mu$ and $\mu'$ agree on the Borel $\sigma$-algebra (or any $\sigma$-algebra with respect to which all functions in $C_c(X)$ are measurable), then for each $f\in C_c(X)$ you have $\int f\,\mathrm{d}\mu=\int f\,\mathrm d\mu'$, even if there is some (non-Borel) set $A$ for which $\mu,\mu'$ are defined and $\mu(A)\neq \mu'(A)$ (although in this case, they can't both be regular).
Further, if $\mu,\mu'$ are distinct Radon Borel measures (that is, locally finite, inner regular with respect to compact sets), then they necessarily disagree about the measure of some compact set $K$, which can be used to construct some $f\in C_c(X)$ which has different integrals with respect to $\mu,\mu'$.
This can be used to obtain the following statement of the Riesz representation theorem:

If $X$ is a locally compact Hausdorff space, then the space $C_c(X)^*$ (the continuous dual) is naturally isomorphic (as a topological vector space) to the space of all Radon signed (or complex) Borel measures on $X$ (with the corresponding weak topology).

The Borel $\sigma$-algebra can be replaced by the $\sigma$-algebra generated by the preimages of open intervals by continuous functions (which may be strictly smaller), and usually also by a strictly larger one (consisting of the universally measurable sets); this is all equivalent by the regularity hypothesis. You can also replace $C_c(X)$ by its completion $C_0(X)$
