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Let $\mu $ be the Lebesgue measure on $\mathbb{R}^n$, let $f\in L^1_{loc}(\mathbb{R}^n)$ such that $f\ge 0$

prove the induced measure $\mu _f(E) = \int_Efd\mu $ is regular measure that is:

$$\mu_f(E) = \inf\{\mu_f(U)\mid E\subset U,\text{U is open}\}$$

for any $E$ as Borel set and

$$\mu_f(E) =\sup\{\mu_f(K)\mid K\subset E,\text{K is compact}\}$$

for any open set $E$.

I have 2 idea seems possible,one is since $\mu_f$ defines a positive linear functional on $C_c(\mathbb{R}^n)$ using Riesz representation exist some Radon measure $\nu$ such that $$\int\phi d\nu = \int \phi fd\mu$$

For any $\phi\in C_c(\mathbb{R}^n)$,then may be we can claim $d\nu = fd\mu$ so the induced measure is Radon?

The second approach is using regularility of Lebesgue measure,but the convergence condition is hard to check .

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  • $\begingroup$ Approximate $f$ by a sequence of simple functions and use the monotone convergence theorem to show that $\mu_f$ is inner regular. $\endgroup$
    – user140541
    Commented Dec 8, 2020 at 15:58
  • $\begingroup$ @d.k.o. thanks I got it, do you mean construct a sequence of monotone increasing simple function ,then each $\chi_E$ (in the finite sum of simple function) can be approximate by a compact set $\chi_K$ ? $\endgroup$
    – yi li
    Commented Dec 8, 2020 at 16:05
  • $\begingroup$ Yes. For indicators, $\chi_B$, the (inner) regularity follows from that of $\mu$. $\endgroup$
    – user140541
    Commented Dec 8, 2020 at 16:23

1 Answer 1

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$\mathbb{R}^n$ is second countable LCH space. So, every open set is $\sigma$-compact. Therefore a Borel measure on $\mathbb{R}^n$ that is finite on compact sets is regular and thus Radon (Folland Theorem 7.8). $\mu_f$ is clearly Borel. Since $f \in L^1_{\text{loc}}$, it is finite on compact sets.

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