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?

  • 3
    $\begingroup$ Where did you get that statement of the Riesz theorem? Normally, the measures you get are Borel measures, not measures on some abstract sigma algebra. It is easy to check that regularity is preserved. $\endgroup$
    – tomasz
    Aug 12, 2021 at 16:57
  • $\begingroup$ Real and Complex analysis. Walter Rudin theorem 2.14 $\endgroup$ Aug 12, 2021 at 20:16

1 Answer 1


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 $C_0(X)$


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