This may be an obvious question, but I'm still doubting if my reasoning is ok.
There is a result that says the following:
If $1 \leq p \leq s \leq \infty$, and $|\Omega| < \infty$, ($\Omega$ is an open of $\mathbb{R}^n$), then $L^s(\Omega) \subset L^p(\Omega)$, where
$$L^q(\Omega) := \left\{f : \Omega \to \mathbb{R} \; : \; \int_{\Omega} |f|^q < \infty \right\}$$
if $1 \leq q < \infty$ and
$$L^\infty(\Omega) := \left\{f : \Omega \to \mathbb{R} \; : \; \text{there is } c\in \mathbb{R}_{\geq 0} \text{ such as } |f(x)| < c \text{ for almost all } x \in \Omega\right\}.$$
So with the help of this proposition I think we can say that
$$L^\infty_{loc}(\mathbb{R}^n) \subset L^q_{loc}(\mathbb{R}^n) $$
where $loc$ just means that, $f \in L^q_{loc}(\mathbb{R}^n) $ iff for every compact $K \subset \mathbb{R}^n$, we have that $f|_K \in L^q_{loc}(K)$ for $1\leq q \leq \infty$.
If we take $f \in L^\infty_{loc}(\mathbb{R}^n)$, then for any $K$ compact, we have that $f \in L^\infty(K)$, and thus $f \in L^1(K)$.
Is this reasoning ok?, or I am missing something?
Sorry if this is a duped question.