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In the book measure theory and fine properties of functions of Evans and Gariepy, i'm trying to understand the proof of theorem 2.2 (ii) which states that $\mathcal{L}^1 = \mathcal{H}^1$. Here the Hausdorff measure is defined with the normalized constant, so that $$\mathcal{H}_\delta^1 = \inf \bigg\{ \sum_{j=1}^\infty diam C_j: A \subset \bigcup_{j=1}^\infty C_j, diam C_j \leq \delta\bigg\} $$

I understand the first part of the proof which proves that $\mathcal{L}^1 \leq \mathcal{H}^1(A)$ so I will just type the second one.

Let $\delta > 0$ and $C_j$ a covering $A$ such that $diam C_j \leq \delta$. Consider $I_j = [k\delta,(k+1)\delta]$, then $diam C_j \cap I_k \leq \delta$ and $$\sum_{k=1}^\infty diam(C_j \cap I_k) \leq diam (C_j) $$

Why is this inequality true? I tried to prove it but with no success. Then the proof follows with

$$\mathcal{L}^1 = \inf\bigg\{ \sum_{j=1}^\infty Diam C_j :A \subset \bigcup_{j=1}^\infty C_j\bigg\} \geq \inf\bigg\{ \sum_{j=1}^\infty \sum_{k=1}^\infty Diam C_j \cap I_k : A \subset \bigcup_{j=1}^\infty C_j \bigg\} \geq \mathcal{H}_\delta^1 $$

I don't understand the last inequality, where does it come from? Thanks in advanced!

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Hints: For the first question use the fact that if $E$ is a bounded set of real numbers then the diameter of $E$ equals the Lebesgue measure of the interval from $\inf E$ to $\sup E$.

For the second question use the fact that $A \subset \bigcup_{j=1}^\infty C_j$ implies $A \subset \bigcup_{j,k=1}^\infty (C_j\cap I_k)$ (and use the definition of $\mathcal H^{1}_\delta$

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  • $\begingroup$ Thanks, I already figured out the second part. For the first part I did this: Let $a_k = \inf C_j \cap I_k$ and $b_k = \sup C_j \cap I_k$ and $a = \inf C_j$ and $b = \sup C_j$ then $a \leq a_k$ and $b_k \leq b$ so that $(a_k,b_k) \subset (a,b)$ for all k, and since the sets $(a_k,b_k)$ are pairwise disjoint we have $$\sum_{k=-\infty}^\infty diam C_j \cap I_k =\sum_{k=-\infty}^\infty \mathcal{L}^1(a_k,b_k) = \mathcal{L}^1 \bigg(\bigcup_{k=-\infty}^\infty (a_k,b_k) \bigg) \leq \mathcal{L}^1(a,b) = diam C_j $$ is it correct? $\endgroup$
    – Franlezana
    Commented Jan 30, 2023 at 20:30
  • $\begingroup$ @Franlezana Yes, it is correct. $\endgroup$ Commented Jan 30, 2023 at 23:13

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