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Let $A \subset \mathbb{R}$ such that $\forall x \in \mathbb{R}$ and $r>0$ holds $|A \cap [x,x+r]| \leq rf(r)$ where $\lim_{r \rightarrow 0} f(r) = 0$. Show that $|A| = 0$.

I tried to prove the statement when $A$ is bounded, but i failed. Therefore any help is apperciated, but rather a hint than a solution. I guess the general case follows from the bounded case.

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    We have for $N\in \mathbb{N}{\geq 1}$ $$ \vert A\cap [0,1]\vert = \sum{j=1}^N \vert A\cap [(j-1)/N, j/N]\vert \leq \sum_{j=1}^N N^{-1} f(N^{-1}).$$ I guess you can take from here. – Severin Schraven Mar 06 '21 at 14:34

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Suppose that $A\subset [u,v]$, for every $n>0$, denote $A_{i,n}=[u+{i\over n}(v-u),u+{{i+1}\over n}(v-u)]\cap A$, $A\subset \cup_{i=0}^{i=n-1}A_{i,n}$, $\mu(A_{i,n})\leq {{u-v}\over n}f({{u-v}\over n})$.

This implies that $\mu(A)\leq\sum_{i=1}^{i=n-1}{{v-u}\over n}f({{v-u}\over n})$.

For every $c>0$, there exists $N$ such that for every $n>N$, $f({{v-u}\over n})<c$, this implies that $\mu(A)\leq \sum_{i=1}^{i=n-1}{{v-u}\over n}f({{v-u}\over n})\leq \sum_{i=1}^{i=n-1}c{{v-u}\over n}\leq c(v-u)$. This implies that $\mu(A)=0$/

Tsemo Aristide
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