I am studying real analysis and I have some problems in understanding properties of Hausdorff measure.

Let $\mathcal{E}_\delta$ be collection of subsets of $\mathbb{R}^N$ whose diameter is less then $\delta$. Now $\mathcal{E}_\delta$ has the property that for every $E \subseteq \mathbb{R}^N$ there exists a countable collection $\{ Q_n\}_{n \in \mathbb{N}} \subseteq \mathcal{E}_\delta$ such that $E \subseteq \bigcup_{n \in \mathbb{N}}Q_n$ (to see this we may observe that $\mathbb{R}^N$ could be partitioned in a numerable union of dyadic cubes, each with diameter arbitrarily small).

Now define $$ \mathcal{H}_{\alpha,\delta}(E)=\inf \left\{ \sum_{n \in \mathbb{N}} \text{diam}(E_n)^\alpha \,\middle|\, E \subseteq \bigcup_{n \in \mathbb{N}} \ \text{and} \ \left\{E_n \right\} \subseteq \mathcal{E}_\delta \right\} $$ By the properties of infimum we deduce that if $\delta_1 \le \delta_2$ then $\mathcal{H}_{\alpha,\delta_2}(E) \le \mathcal{H}_{\alpha,\delta_1}(E)$. So it make sense define the Hausdorff outer $\alpha$-measure of $E$ by $$\mathcal{H}_\alpha(E)=\sup_{\delta>0} \mathcal{H}_{\alpha,\delta}=\lim_{\delta \to 0^+}\mathcal{H}_{\alpha,\delta} $$ Now $\mathcal{H}_\alpha$ is an outer measure, and if consider the collection $\mathcal{A}$ of sets $E \subseteq \mathbb{R}^N$ such that $$\mathcal{H}_\alpha(A)=\mathcal{H}_\alpha(E \cap A) + \mathcal{H}_\alpha(A\setminus E), \ \ \forall A \subseteq \mathbb{R}^N \ \ \ \ \ (*) $$ we can prove (using the Carathéodory Extension Theorem) that $\mathcal{A}$ is a $\sigma$-algebra, and we say that $\mathcal{A}$ is the collection of Hausdorff $\alpha$-measurable sets.

Now I have two questions:

  • There are some difference between $\mathcal{A}$ and the collection of Lebesgue measurable sets?
  • There is some subset of $\mathbb{R}^N$ that is not Hausdorff $\alpha$-measurable?

For answering the second question I taken $(*)$, that for sub-additivity of outer measure holds ever with the sign $\le$ at place of $=$ and I looked for a subset $E$ of $\mathbb{R}^N$ for that exists a subset $A$ of $\mathbb{R}^N$ such that $(*)$ holds with strict inequality. For the candidate of $E$ I have taken the Vitali set $V$, and for the set $A$ I take the set $$A= \bigcup_{q \in (-1,1) \cup \mathbb{Q}} V+q$$ where $V+q$ is the translated of Vitali set (like the proof that $V$ is not Lebesgue measurable). I easy to see that $V+q_1 \cap V+q_2=\emptyset$ if $q_1 \neq q_2$. Now what remain to check is that holds the strict inequality $$ \mathcal{H}_{1}(A)<\mathcal{H}_1(V)+\mathcal{H}_1 \left(A \setminus V\right)$$

Where $$A \setminus V =\bigcup_{q \in(-1,1) \cap \mathbb{Q}\setminus \{0\} } V+q $$ Now $V \subseteq (0,1)$, then $A \subseteq (-1,2)$ so for the sub-additivity $\mathcal{H}_1(V) \le 1$ and $\mathcal{H}_1(A) \le 3$, and if we can show that $\mathcal{H}_1(A \setminus V)=\mathcal{H}_1(A)$ and $\mathcal{H}_1(V)>0$ we are done. But if the above reasoning were false and the Vitali set were Hausdorff $1$-measurable, then the first question would have answer.


In $\mathbb R^N$, the Lebesgue measurable sets are the same as the sets measurable for $N$-dimensional Hausdorff measure. But not the same as the sets measurable for $\alpha$-dimensional Hausdorff measure when $0<\alpha < N$.

For a non-measurable set (when $\alpha \ne N$) I would use the Bernstein set rather than the Vitali set.


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