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The answer is no. Let $m$ be a very large positive constant. You can find smooth $f$ that equals $m$ on a small ball $B_R(0)$ and still satisfy $\int_{B_6(0)}|f|<\delta$. You can do it with $R$ such t …

Let $s>1$ be aby number. The unit interval $[0,1]$ with the metric $d(x,y)=|x-y|^{1/s}$ has poisitive and finite $s$-dimensional Hausdorff measure. While, it is an abstract metric space, it can be emb …

$w\in \partial B(z,ts)$, so $|w-z|=ts$ and hence $$ \frac{s^p}{t^{n-1}}=s^{n+p-1}|w-z|^{1-n}. $$ You can also check Lemma 2.33 in my Lecture notes, where the proof of the result you are studying is pr …

If $s>1$, then clearly $H^s(B_x)=0$ so there is nothing to do. If $s=1$, $H^1$ is just the Lebesgue measure so measurability follows. If $0<s<1$ the situation is a way more complicated, but the answer …

The next result answers the question in the negative. Theorem. There is $\phi:\mathbb{R}^n\supset\Omega\to\mathbb{R}^n$ of class $C^\infty$ such that $\phi$ is a local diffeomorphism in a neighborhoo …

Unfortunately, you can always find a Lipschitz map $$ f:\mathbb{R}^m\to\mathbb{R}^{m-k} \quad \text{and} \quad y\in\mathbb{R}^{m-k} $$ such that $\partial f^{-1}(y)$ has positive $m$-dimensional measu …