TITLE: Lebesgue Measure: No Atoms!
QUESTION [7 upvotes]: Disclaimer: This is just meant as record of a proof. For more details see: Answer own Question

How to prove that the Lebesgue measure has no atoms:
$$\lambda:\mathbb{R}^n\to\mathbb{R}_+$$
(Recall the definition for atoms: Measure Atoms: Definition?)

REPLY [8 votes]: Here is one way: Let $B_r$ be the box $\{x\in\mathbb{R}^n: |x_k|\le r\text{ for } k=1,\ldots,n\}$. You can easily prove that $\lambda(B_r\cap A)$ is a continuous function of $r$, for any Lebesgue measurable set $A$. From this, it follows immediately that $A$ is not an atom.
Edit: To explain the last step, $\lambda(B_r\cap A)$ is a continuous function of $r$ taking the value $0$ at $r=0$ and $\lambda(A)$ in the limit $r\to\infty$. But if $A$ is an atom, this function can only take two values.