$\sum_{i\in\mathbb{N}}\text{diam}(A_i)\geq \text{diam}(I)$ when $I\subset \bigcup_{i\in \mathbb{N}} A_i$?

Question

I came across with this kind of problem when considering Hausdorff measures. If I have an interval, let's say $I=[a, b]$, and sets $(A_i)_{i\in\mathbb{N}}$ s.t. $I\subset \bigcup_{i\in \mathbb{N}} A_i$, does it hold that $\sum_{i\in\mathbb{N}}\text{diam}(A_i)\geq \text{diam}(I)$? This seems plausible when drawing a picture but I'm not able to do it rigorously. Does this hold for general sets in metric spaces? My motivation for this problem is to show that $\mathcal{H}^1_{\delta}(I)=b-a$.

Answer 1

I think it's true. Here are two steps to reduce it to the case of open intervals:

1. may replace $A_i$ with a small $\epsilon_i$ open neighborhood

2. every open subset $U$ is a countable disjoint union of intervals, and $\operatorname{diam}(U) \ge \sum \operatorname{diam}(U_n)$