Why a map in $W_{{loc}}^{1,\frac{n}{2}}$ whose differential is in $L_{loc}^{n}$ belongs to $W_{{loc}}^{1,n}$?

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Let $\Omega \subseteq \mathbb{R}^n$ an open subset. Let $f \in W_{{loc}}^{1,\frac{n}{2}}(\Omega,\mathbb{R}^n)$, and suppose also that $df \in L_{loc}^{n}(\Omega,\mathbb{R}^{n^2})$.

How can I prove that $f \in W_{{loc}}^{1,n}(\Omega,\mathbb{R}^n)$?

Clearly $|df|^n$ is locally integrable. We need to prove $|f|^n$ is locally integrable.

My first thought was to use Sobolev inequality:

Let $p<n$. For every $f \in W^{1,p}_0(\Omega,\mathbb{R}^n)$ $$ \|f\|_{p^*} \le C \| df\|_p \tag{1}$$ where $p^*$ is the Sobolev conjugate of $p$, i.e. $1/p=1/p^*+1/n$. Taking $p=n/2$ we get $p^*=n$, so inequality $(1)$ becomes

$$ \|f\|_{n} \le C \| df\|_{n/2}. \tag{2}$$

The problem is that Sobolev inequality only holds for maps in $W^{1,p}_0$, not in $W^{1,p}$ in general. (Unless $\Omega$ satisfies some very specific properties, which I do not want to assume).

Am I missing something obvious?

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The goal is to prove that $|f|^n$ is locally integrable. This is equivalent to being integrable on every ball $B$ contained in $\Omega$. The Sobolev embedding theorem applies on $B$, as it "satisfies some very specific properties" which we do not need to assume of $\Omega$ itself.