Let $\Omega$ be an open set in $R^n$, and $f \in L^1_{loc}(\Omega)$, such that foreach multiindex $\alpha\in N^n$, $|\alpha| = l$ f has weak derivative $D^\alpha f$ in $L^p(\Omega)$, with $1\leq p\leq \infty$.

In general it is not true that $f\in L^p(\Omega)$, but it have to be true that $f\in L^p_{loc}(\Omega)$. How can be shown that?

Let $\Omega$ be an open set in $R^n$, and $f \in L^1_{loc}(\Omega)$, such that for each multiindex $\alpha\in N^n$, $|\alpha| = l$ f has weak derivative $D^\alpha f$ in $L^p(\Omega)$, with $1\leq p\leq \infty$.

In general it is not true that $f\in L^p(\Omega)$, but it has to be true that $f\in L^p_{loc}(\Omega)$. How can this be shown?

Let $\Omega$ be an open set in $R^n$, and $f \in L^1_{loc}(\Omega)$, such that foreach multiindex $\alpha\in N^n$, $|\alpha| = l$ it holds f has weak derivative $D^\alpha f$ in $L^p(\Omega)$, with $1\leq p\leq \infty$.

In general it is not true that $f\in L^p(\Omega)$, but it have to be true that $f\in L^p_{loc}(\Omega)$. How can be shown that?

Let $\Omega$ be an open set in $R^n$, and $f \in L^1_{loc}(\Omega)$, such that foreach multiindex $\alpha\in N^n$, $|\alpha| = l$ f has weak derivative $D^\alpha f$ in $L^p(\Omega)$, with $1\leq p\leq \infty$.

In general it is not true that $f\in L^p(\Omega)$, but it have to be true that $f\in L^p_{loc}(\Omega)$. How can be shown that?