Let $\mu$ be an outer measure on $\mathbb{R}$ such that $\mu([\alpha, \beta])\leq(\beta-\alpha)^2\quad\forall-\infty<\alpha<\beta<\infty$. Prove that $\mu=0$.
I'm thinking about using notable series like $\sum_{n\in\mathbb{N}}\frac{1}{n^2}$ but I don't understand. Can you help me?
Edit: General version. Let $\mu$ be an outer measure on $\mathbb{R}^n$, $n\geq 2$, such that $$\mu(B_r(x))\leq r^{n+\epsilon}$$ for all $r>0$, $x\in\mathbb{R}$ and for a fixed $\epsilon>0$. Prove that $\mu=0$.
Notation: $B_r(x)=\{y\in\mathbb{R}^n:|y-x|<r\}$ is the Euclidean ball.
Here is my solution:
Fix some finite interval $[a,b]\subseteq\mathbb{R}$ and pick an arbitrary integer $n\geq1$. Since $$[a,b]=\bigcup_{k=0}^{n-1} \left[a+\frac{k(b-a)}{n},a+\frac{(k+1)(b-a)}{n}\right],$$ we can use the subadditivity of $\mu$ to see that $$\mu([a,b])\leq\sum_{k=0}^{n-1}\mu\left(\left[a+\frac{k(b-a)}{n},a+\frac{(k+1)(b-a)}{n}\right]\right)\leq\sum_{k=0}^{n-1}\left(\frac{b-a}{n}\right)^2=\frac{(b-a)^2}{n}.$$ Since $n$ was arbitrary, it follows that $\mu([a,b])=0$. This implies that $\mu=0$ (since every subset of $\mathbb{R}$ can be covered by a union of finite intervals).
Edit: It is possible to use a similar idea in the general case. The details are as follows. First, observe that if $C$ is a cube in $\mathbb{R}^n$ with sidelength $s$, then $C$ is contained in a ball of radius $n^{1/2}s$, so $\mu(C)\leq \alpha s^{n+\epsilon}$ where $\alpha=n^{(n+\epsilon)/2}$. Let $A$ be an arbitrary cube in $\mathbb{R}^n$ with sidelength $r$. We will show that $\mu(A)=0$. Fix any integer $k\geq1$, and notice that we can find a collection of $k^n$ cubes with side-length $r/k$ such that their union equals $A$. Now, we can use the subadditivity of $\mu$ to see that $$\mu(A)\leq k^n\alpha \left(\frac{r}{k}\right)^{n+\epsilon}=\frac{1}{k^{\epsilon}}\alpha r^{n+\epsilon}.$$ Since $k$ is arbitrary, this implies that $\mu(A)=0$. Finally, as $\mathbb{R}^n$ can be written as a countable union of cubes, it now follows that $\mu=0$.