I wish to show the following statement:
Nonnegative $f \in L^1_{loc}(\mathbf{R}^n, \mathcal{B}_{\mathbf{R}^n}, \lambda)$ if and only if its induced Borel measure $\mu_f(E) := \int_{E} f \,d\lambda$ is locally finite and regular (inner-regular and outer-regular for all Borel sets).
I have proven and sure of my proof for showing locally finiteness and outer-regularity. I would like to verify my proof for the inner-regularity of $\mu_f$ for any given $E \in \mathcal{B}_{\mathbf{R}^n}$:
Suppose first that the set $E$ is bounded and thus have finite Lebesgue measure i.e. $\lambda(E) < +\infty$. Now note the Lebesgue measure $\lambda$ is inner-regular and thus for all $\delta > 0$, there exists $K \subseteq E \subseteq \mathbf{R}^n$ compact such that $\lambda(E - K) < \delta$. Now observe $\mu_f$ is by definition absolutely continuous with respect to $\lambda$, for any given $\epsilon > 0$, there exists $\delta_\epsilon > 0$ such that $\lambda(E) < \delta_\epsilon \implies \mu_f(E) < \epsilon$. Therefore, we have for any $\epsilon > 0$, there exists $K \subseteq E$ such that $\lambda(E - K) < \delta_\epsilon \implies \mu_f(E) - \mu_f(K) = \mu_f(E - K) < \epsilon$. That is, $\forall \epsilon > 0$, $\exists K \subseteq E$ such that $\mu_f(E) - \epsilon < \mu_f(K)$. This shows $\mu_f$ is inner regular on bounded $E$.
Now lift the restriction on the boundedness of $E \in \mathcal{B}_{\mathbf{R}^n}$, for $j \in \mathbf{N}$, define the sets $E_j := E \cap [-j, j]^n \in \mathcal{B}_{\mathbf{R}^n}$ bounded. Then we have by the above proof for bounded sets there exists $K_j \subseteq E_j$ compact with $\mu_f(E_j) - \frac{1}{j} < \mu_f(K_j) \leq \mu_f(E_j)$. Taking the limsup and liminf on both sides, we see as $\lim_{j} \mu_f(E_j) = \mu_f(\bigcup_{j} E_j) = \mu_f(E)$ that $$ \lim_j \mu_f(K_j) = \mu_f(E). $$ Following from the definition of limit (one would discuss this in cases:$\mu_f(E) = +\infty$ or $< +\infty$ in $\epsilon$-$\delta$ definition), we conclude $\mu_f$ is inner-regular on $E \in \mathcal{B}_{\mathbf{R}^n}$ i.e. $\mu_f(E) = \sup\{ \mu_f(K): K \subseteq E \text{ compact} \}$.
Is the proof correct?