A measure is a result of Lebesgue measure Define an infinite measure $\mu$ on $(\mathbb{R}, \mathcal{R})$ (i.e. $\mu (B) < \infty$ with an arbitrary bounded set $B \in \mathcal{R})$. We further assume $\{B + x \} = \{x + y \in \mathbb{R}; y \in B \} $ and let $\mu(B+x) = \mu(B)$.
Prove that $\mu = c \xi$ with $c \geq 0$ and $\xi$ is a Lebesgue measure.
My opinion: If I assume that $\mu$ is a Lebesgue measure on $(\mathbb{R}, \mathcal{R})$, then we always have $\mu(B+x) = \mu(B)$ and $\mu(cB) = c \mu(B)$. On the other hand, the problem is trivial. What do you think about my solution? I seem to be missing something.
 A: I assume the interior of $B$ is non-empty (that is to say $B$ contains an open interval).


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*Since $\mu$ is a measure, $\mu((0,1]) = n\cdot \mu((0,\frac 1n])$ for any $n\in \mathbb N$. Hence $\displaystyle \mu((0,\frac 1n]) = \frac{\mu((0,1])}n$.

*Let $a=\frac pq$, $b=\frac{p'}{q'}$ be rationals with $q,q'>0$ and $a<b$. Then $$\begin{align}\mu((a,b]) &= \mu((0,b-a]) \\&= \mu((0, \frac{p'q-pq'}{qq'}]) \\&= (p'q-pq')\mu((0, \frac{1}{qq'}])\\&=(p'q-pq') \frac{\mu((0,1])}{qq'}\\&= (b-a)\mu((0,1])\end{align}$$

*Let $a\in \mathbb R$. Since $\displaystyle \{a\}=  \cap_n (a-\frac 1n, a]$, you can find $n$ large enough such that a translation of $\displaystyle (a-\frac 1n, a]$ is in the interior of $B$, and you can assume without loss of generality that $\mu((a-\frac 1N,a]) <\infty$, hence $$\mu(\{a\})=\lim_n \mu((a-\frac 1n, a])=\lim_n \mu((0, \frac 1n])=\mu((0,1])\cdot 0 = 0 $$

*Let $a,b\in \mathbb R$. Let $a_n$ be a decreasing sequence of rationals such that $a_n\to a$ and $b_n$ be a increasing sequence of rationals such that $b_n\to b$. Then  $$\begin{align}\mu((a,b]) &= \mu ((a,b)) \\&= \mu(\cup_n(a_n,b_n))\\&=\lim_n \mu((a_n,b_n)) \\&= \lim_n \mu((a_n,b_n])\\&=(b_n-a_n)\mu((0,1]) \\ &= (b-a)\mu((0,1])\end{align}$$

*$\mu$ and $\mu((0,1])\xi$ coincide on a $\pi$-system, continuity of both yields $\mu(\mathbb R) = \mu((0,1])\xi(\mathbb R)$ and $\xi$ is $\sigma$-finite. A well-known lemma proves that $\mu$ and $\mu((0,1])\xi$ coincide on the $\sigma$-algebra generated by the $(a,b]$, hence coincide on the Borel $\sigma$ algebra.

*Let $A$ be Lebesgue measurable. then $A=B\cup C$ where $B$ is Borel and $C$ is a null set w.r.t to $\xi$. You already know $\xi(A)=\xi(B)$. There is some $D$ in the Borel $\sigma$-algebra such that $C\subset D$ and $\xi(D)=0$, hence $\mu(D)=0$. Then $$\mu(B)\leq \mu(B\cup C)=\mu(A)\leq \mu(B\cup D)\leq \mu(B)+\mu(D) = \mu(B) $$
Hence $\mu(A)=\mu(B) = \mu((0,1])\xi(B)=\mu((0,1])\xi(A)$.
$\mu((0,1])\xi$ and $\mu$ coincide on all of the Lebesgue $\sigma$ algebra.
