Positive measure Let $\mu$ positive measure over $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Prove that if $\mu$ is invariant translations and $\mu([0,1])=1$, then $\mu([0,\frac{1}{n}])\leq\frac{1}{n}$
I know that this measure is the Lebesgue measure, but I can´t prove that  $\mu([0,\frac{1}{n}])\leq\frac{1}{n}$
 A: Fix $n\geq 1$ and consider the partition $\left\{[\frac{k-1}{n},\frac{k}{n})\right\}_{k=1}^{n}$ of the half-open interval $[0,1)\in \mathcal{B}(\mathbb{R})$
Now since $\mu$ is a positive measure we have that $$\sum_{k=1}^{n}\mu\left([\frac{k-1}{n},\frac{k}{n})\right)= \mu([0,1)) \leq \mu([0,1]) =1 $$ 
Now since $\mu$ is assumed to be translation invariant, we have that $\mu([0,\frac{1}{n})) = \mu\left([\frac{k-1}{n},\frac{k}{n})\right)$ for all $1<k\leq n$. Hence $$\mu([0,\frac{1}{n})) \leq \frac{1}{n}$$
Now i claim that $\mu(\left\{x\right\})=0$ for any $x \in \mathbb{R}$, since if there would be a singleton with positive measure, then certainly every singleton would have positive measure due to the translation invariance, which contradicts the fact that $\mu([0,1]) =1$ by simply taking any countably infinite collection of singleton subsets of $[0,1]$.
Thus we arrive at $$\mu([0,\frac{1}{n}]) = \mu([0,\frac{1}{n})) + \mu\left(\left\{\frac{1}{n}\right\}\right) = \mu([0,\frac{1}{n})) \leq \frac{1}{n}$$
