# prove measure induced by locally integrable function is Radon measure

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 .

• Approximate $f$ by a sequence of simple functions and use the monotone convergence theorem to show that $\mu_f$ is inner regular.
– user140541
Commented Dec 8, 2020 at 15:58
• @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 $\chi_K$ ? Commented Dec 8, 2020 at 16:05
• Yes. For indicators, $\chi_B$, the (inner) regularity follows from that of $\mu$.
– user140541
Commented Dec 8, 2020 at 16:23

$$\mathbb{R}^n$$ 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.