How to prove that $\mu(\{x\}) = \mu(\{y\}) = 0$ for a measure $\mu$? Sorry for my bad english.
Let $\mu$ a measure on the Borel set. For all interval $I \subset \mathbb{R}$ and all $x \in \mathbb{R}$, we note $I + x = \{y \in \mathbb{R} : \exists z \in I, y = z + x\}$. We suppose $\mu$ verify the following property : $\mu(I) = \mu(I+x)$ for all interval $I$ and all real number $x$, and $\mu([0,1]) = 1$.
We want to show that $\forall (x,y) \in \mathbb{R}^2, \mu(\{x\}) = \mu(\{y\}) = 0$.
I've done : $\mu(\{x\}) = \mu(\{x\}+x)$ with $\{x\} + x = \{m \in \mathbb{R} : \exists z \in \{x\}, m = z + x = 2x\}$.
And $\mu(\{y\}) = \mu(\{y\}+y)$ with $\{y\} + y = \{m \in \mathbb{R} : \exists z \in \{y\}, m = z + y = 2y\}$.
I don't know how to conclute it. I think it's easy, but I'm a beginner in measure theory. And I suppose $\mu$ will be the Lebesgue measure, but I can't prove it... Someone could help me ? Thank you in advance.
 A: Hint You only need to prove that $\mu(\{ a \})=0$ for all $a \in \mathbb R$.
Assume by contradiction that there exists some $a$ so that $\mu(\{a \})=b >0$. 
Set $I=(a-\epsilon, a+\epsilon)$. then 
$$\mu(I) \geq \mu(\{a \})=b$$
Now, for each positive integer $n$ pick $\epsilon=\frac{1}{2n}$ and partition $[0,1]$ into $n$ intervals of lenght $\frac1n$. Use the above together with the invaraince under translates to conclude
$$1=I([0,1]) \geq nb$$
Since this is true for all $n$, this contradicts $b >0$.
A: Once we have established that $\mu(\{x\})=0$ for every real $x,$ we can show that $\mu(S)$ is equal to the Lebesgue measure $\lambda (S)$ for every Borel set $S.$
(1). For $n\in \mathbb N$ we have $$1=\mu([0,1])=\mu((0,1))+\mu(\{0\})+\mu(\{1\})=\mu((0,1))=$$     $$=\sum_{j=0}^{n-1}\mu((j/n,(j+1)/n))+\sum_{0<j<n}\mu(\{j/n\})=$$    $$=\sum_{j=0}^{n-1}\mu((j/n,(j+1)/n))=\sum_{j=0}^{n-1}\mu (j/n+(0,1/n))=\sum_{j=0}^{n-1}\mu((0,1/n))=$$ $$=n\mu((0,1/n)).$$
So $\mu((0,1/n))=1/n.$
And $\mu ((a,a+1/n])=\mu ((0,1/n])=\mu((0,1/n))+\mu(\{1/n\})=1/n$ for any $a$.
(2). For $r>0$ and  $n\in \mathbb N$  let $s(r,n)\in \mathbb N\cup \{0\}$ such that $s(r,n)/n< r\leq(1+s(r,n))/n.$  Then $$\cup_{j=0}^{-1+s(r,n)}(j/n,(j+1)/n]\subset\;(0,r)\;\subset \cup_{j=0}^{s(r,n)} (j/n,(j+1)/n].$$  So $s(r,n)/n\leq \mu ((0,r))\leq (1+s(r,n))/n.$  Since $\lim_{n\to \infty}s(r,n)/n=r,$ we have $\mu((0,r))=r. $
So for $a<b$ we have $\mu((a,b))=\mu((0,b-a))=b-a.$
(3). So $\mu$ agrees with Lebesgue measure $\lambda$ on all bounded open intervals, and $\mu$ is zero on finite sets. So $\mu$ agrees with $\lambda$ on all open sets.  And $\mu$ is translation-invariant on all Borel sets. Therefore  $\mu(S)=\lambda(S)$ for every Borel set $S.$ 
