Theorem. Let $\mu(X)<\infty$. Then $$1\le p \le q\le\infty\implies L^q(X,\mathcal{A},\mu)\subseteq L^p(X,\mathcal{A},\mu) $$
Def. Let $\Omega\subseteq\mathbb{R}^n$ be an open subset, $f\colon\Omega\to [-\infty,+\infty]$ q.o defined. We said that the function $f$ is locally integrable in $\Omega$ if $f\in L^1(G,\mathcal{L}(\mathbb{R}^n)\cap G,\lambda)$ for all $G\in\mathcal{L}(\mathbb{R^n})$ with compact closure $\overline{G}\subseteq\Omega.$
In the definition $\lambda$ is the Lebesgue measure on $\mathbb{R}^n$.
We denote the set of locally integrable function with $L^1_{\text{loc}}$
I must prove that $$L^p(\Omega)\subseteq L^1_{\text{loc}}(\Omega)\quad\text{for all}\;p\in[1,+\infty]$$
using the previous theorem.
Naturally $$L^1(\Omega)\subseteq L^1_{\text{loc}}(\Omega)$$
Now I don't know how to proceed. Could anyone give me a suggestion? Thanks!
Best Answer
Take $K\subset \Omega $ compact then $\lambda(K)<\infty$. Therefore, $(K,\mathcal{L}(\mathbb{R}^n)\cap K,\lambda)$ is a measure space with finite measure. Then take $f\in L^p(\Omega)$, since $$\nu(E)=\int_E |f|^pd\lambda$$ is a measure and $K\subset \Omega$ then $$\int_K|f|^pd\lambda\leq \int_{\Omega}|f|^p d\mu<\infty.$$ Therefore, $f\in L^p_{loc}$. Finally, since $(K,\mathcal{L}\cap K,\lambda)$ is a measure space with finite measure, $1\leq p$ and $f\in L_{loc}^p$, using your theorem with $q=1$ then $f\in L_{loc}^1$. So you get the inclusion $$L_p\subset L_{loc}^1$$