# Proof clarification of hausdorff measure equals lebesgue measure in one dimension

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!

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 $$\mathcal H^{1}_\delta$$
• 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? Commented Jan 30, 2023 at 20:30