Riesz's theorem on representation of positive linear functional on locally compact space as stated in Rudin's "Real and Complex Analysis" assures us that certain $\sigma-$algebra containing all Borel sets exists together with appropriate measure that represents given functional. If we restrict this measure to Borel sets all Borel functions remain measurable. Are their integrals w.r.t. this restricted measure equal to integrals w.r.t. original measure?


Yes. Let $f$ be non-negative and integrable with respect to a measure $\mu$ that is inner regular on sets of finite measure (which exists by Riesz). Let $\epsilon >0$ be given, then there is a simple function $\phi = \sum_1^n a_k \chi_{A_k}$ with $0\leq \phi \leq f$ such that $\int f d\mu - \frac{\epsilon}{2} < \sum_1^n a_j \mu(A_j)$, and $\mu(A_j) < \infty$. By inner regularity, there is a compact set $K_j \subset A_j$ such that $\mu(A_j) - \frac{\epsilon}{2na_j} < \mu(K_j)$. Thus

$$\int f d\mu - \frac{\epsilon}{2} < \sum_1^n a_j \mu(K_j) + \frac{\epsilon}{2},$$ and $$ \int f d\mu - \epsilon < \sum_1^n a_j \mu(K_j) = \int \sum_1^n a_j \chi_{K_j} d\mu.$$

This shows $$\int f d\mu = \sup \left\{ \int \phi d\mu: \phi \text{ is a Borel simple function and } 0 \leq \phi \leq f \right\},$$

which implies the result because it shows we can approximate the integral of any non-negative, $\mu$-measurable function from below with simple functions with compact support. And since compact sets are Borel, this implies we can approximate from below with only Borel sets. Thus the integral will be unchanged if we restrict the measure to only Borel sets.

  • $\begingroup$ Thanks. Now I understand the relevance of regularity properties of borel measures :-) $\endgroup$ – Blazej Jul 21 '15 at 18:14

I believe so. You want to consider the integral $\int f \,d\mu$ of some Borel measurable function. Since such an integral is a limit of integrals of simple functions, it suffices to show that the measure of a Borel set remains unchanged under this restriction. But if $E$ is a Borel set, then by the outer-regularity, $$ \mu(E) = \inf\{\mu(V): V\supset E, V \text{ open}\}. $$ Since all open sets are Borel, they remain in the restricted measure's $\sigma$-algebra. Hence, this infimum does not change when we restrict the $\sigma$-algebra, so neither does $\mu(E)$. Hence characteristic functions of Borel sets have their integrals unchanged, and the claim follows.

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    $\begingroup$ The value of μ(E) shouldn't change because we are making a restriction. You don't need outer regularity to conclude that. What's happening is the set of approximating simple functions is now smaller, so you have to argue that the value of the integral can be obtained with only Borel simple functions. $\endgroup$ – Philip Hoskins Jul 21 '15 at 16:08

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