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 .