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This is a partial answer. In $\mathbb{R}^n$, $\mathcal{H}^n$ coincides with the Lebesgue measure so $\mu(\mathcal{H}^n)$ coincides with the $\sigma$-algebra of Lebesgue measurable sets.

The answer is yes. In what follows I will refer to the book [EG]. By Example on p.198 we know wthat if $U\subset\mathbb{R}^n$ is open with smooth boundary and $\mathcal{H}^{n-1}(\partial U)<\infty$, t …

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 …

A Borel measurable function $f:\mathbb{R}^d\to\mathbb{R}^d$ such that $f(x)\in\partial\varphi(x)$ for all $x\in\mathbb{R}^d$ exists. This follows from the beautiful and surprisingly unknown Federer- …

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 …

While I am still looking for references, let me sketch a proof. This will be a brief sketch only. Theorem. If $f:\mathbb{R}^n\to\mathbb{R}^m$ is Lipschitz, differentiable at $x_o$ and $\operatorname{ …

The answer is no and the following result provides a quite interesting counterexample. This is a known result, but I am not sure where to find it in the literature. Theorem. If $f\in L^1_{\rm loc}(\m …

It is false without assuming that the measure is $\sigma$-finite. Let $\mu$ be the counting measure on $\mathbb{R}$. Then $h>0$ a.e. means $h>0$ everywhere and clearly $$ \int_{\mathbb{R}}hd\mu=\sum_{ …

The following argument is used in my recent paper with D. Azagra and A. Cappello (work in progress). The answer is yes. I asked about whether the proof of Theorem 1 can be modified so to cover Theorem …

Take a set $K\subset\mathbb{R}^2$ that is homeomorphic to the Cantor set and has positive $2$-dimensional Lebesgue measure. Take a homeomorphism of the ternary Canor set onto $K$ and extend it as a pi …

I have an example which does not strictly answer the question, but a slightly weaker one. Theorem 1. There is a function $f:\mathbb{R}^n\to\mathbb{R}$ that is approximately Fréchet differentiable alm …

There is a much more general result of Vallée-Poussin from which a negative answer to your question follows. Let $(X,\mu)$ be a measure space. We say that a family of function $\mathcal{F}\subset L^1( …

I will mention seven different applications: Characterization of almost everywhere differentiability. The following result is a consequence of the Rademacher theorem: Theorem (Stepanov). A functi …

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 answer is yes. I wrote a proof using YCor's comment. Theorem. There a finitely additive measure defined on all subsets of positive integers $\mathbb{N}$, with values into $\{0,1\}$, (only two …