# 英国代考|MATH0017 Measure Theory 测量理论

Proof Throughout the proof we use the fact that $\overline{\mathcal{H}}^{\phi} \mid \mathscr{B}(X)=\mathcal{H}^{\phi}$. We can assume that $\overline{\mathcal{H}}^{\phi}(A)<\infty$, otherwise $G=X$ would do the job. Fix $k \in \mathbb{N}$ and $\eta>0$. Because of Lemma $18.9$, there is a $1 / k$-cover of open sets $\left(U_{i}^{k}\right){i \in \mathbb{N}}$ such that $$\overline{\mathcal{H}}{1 / k}^{\phi}(A) \leqslant \sum_{i=1}^{\infty} C\left(U_{i}^{k}\right) \leqslant \overline{\mathcal{H}}{1 / k}^{\phi}(A)+\eta$$ Since the sets $U^{k}:=\bigcup{i=1}^{\infty} U_{i}^{k}$ are open, $G:=\bigcap_{k \in \mathbb{N}} U^{k}$ is a $G_{\delta}$-set. Monotonicity and $\sigma$-subadditivity yield

$$\mathcal{H}{1 / k}^{\phi}(G) \leqslant \mathcal{H}{1 / k}^{\phi}\left(U^{k}\right) \leqslant \sum_{i=1}^{\infty} \mathcal{H}{1 / k}^{\phi}\left(U{i}^{k}\right) \leqslant \sum_{i=1}^{\infty} C\left(U_{i}^{k}\right) \leqslant \overline{\mathcal{H}}_{1 / k}^{\phi}(A)+\eta$$
Letting $k \rightarrow \infty$ and then $\eta \rightarrow 0$ shows that $\mathcal{H}^{\phi}(G) \leqslant \overline{\mathcal{H}}^{\phi}(A)$. Since $G \supset A$, the other inequality is trivial.

It is possible to show, using arguments similar to those of Appendix $\mathrm{H}$, that any Borel set with $\mathcal{H}^{\phi}(A)<\infty$ contains an $F_{\sigma}$-set $F$ (i.e. a countable union of closed sets) with $\overline{\mathcal{H}}^{\phi}(A)=\mathcal{H}^{\phi}(F)$, see e.g. [40, Chapter 1, Theorem 22] or Problem 18.3.

If $X=\mathbb{R}^{n}$, we can use ‘nets’ to define Hausdorff measure. Consider the lattice $2^{-r} \mathbb{Z}^{n}, r \in \mathbb{N}$, and denote in all lattices by $Q$ the family of right-open cells $Q(r, \vec{z}):=2^{-r}\left([0,1)^{n}+\vec{z}\right), \vec{z} \in \mathbb{Z}^{n}, r \in \mathbb{N} .$

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