$\mu$ is a Radon measure$\implies \lambda(E):=\int_E f\,d\mu$ is also a Radon measure? Let $(X,B,\mu)$ be a measure space (with $B$ the Borel sigma-algebra) and let Let $f\in L^1(X,\mu)$. I was wondering if

$\mu$ is a Radon measure$\implies \lambda(E):=\int_E f\,d\mu$ is also a
Radon measure

(Should I add more conditions on $f$? Is $f\in L^1$ necessary?).
For instance I would like to prove inner regularity on open sets.
Since $\mu$ is Radon so inner regular on open sets we have, $\mu(O)=\sup\{\mu(K):K\subset O\text{ is compact}\}$ so if I fix $\varepsilon>0$ (that can be arbitrary small), there is $K\subset O$ which is compact and such that $\mu(K)\ge \mu(O)-\varepsilon$. So $$\int_O f\,d\mu=\int_K f\,d\mu + \underbrace{\int_{O\setminus K} f \, d\mu}_{\le|f|_{L^1}\cdot\varepsilon}.$$ I think I can use the same reasonning to show that $\lambda$ is outer regular.
Also, for $K$ compact I have $\lambda(K)=\int_K f\,d\mu \le |f|_{L^1}\mu(K)<\infty$ since $f$ is integrable (so $|f|_{L^1}<\infty)$ and $\mu$ is a radon measure (so $\mu(K)<\infty$). Is this reasoning correct or is  there a quicker way to notice that $\lambda$ is Radon?

Try working with sequences.  Find an increasing sequence of compact sets $K_1 \subset K_2 \subset \dots \subset O$ such that $\mu(O \setminus K_n) \to 0$.  Verify that $1_{O \setminus K_n} \to 0$, $\mu$-almost everywhere.    Now use the dominated convergence theorem to conclude that $\lambda(O \setminus K_n) = \int_{O \setminus K_n} f\,d\mu \to 0$ as well.
