Let $(\mathbb{R}, B_{\mathbb{R}}, \lambda)$, where $B_{\mathbb{R}}$ is the Borel $\sigma$-algebra of $\mathbb{R}$ and $\lambda$ is the measure such that $$\lambda((0,1])=1 \quad \mbox{and} \quad \lambda(B+t)=\lambda(B),$$ for all $B \in B_{\mathbb{R}}$ and $t \in \mathbb{R}$. Show that $$\lambda(B)=m(B)$$ for all $B \in B_{\mathbb{R}}$, where $m$ is the Lebesgue measure.
I didn't get this exercise. I''ll apreciatte your answer or tip about it. Thanks in advanced.
Note first that, for any $x,y\in\mathbb R$, we have $\lambda(\{y\})=\lambda(\{y\}x-y)=\lambda(\{x\})$. Then $\lambda(\{x\})=0$ for all $x$. If that were not the case, we would have $$ \infty=\lambda\Big(\bigcup_n\{\tfrac1n\}\Big)\leq\lambda((0,1])=1, $$ a contradiction. In particular $\lambda([0,1])=\lambda((0,1])$, so we don't have to be careful about endpoints.
We have $$ 1=\lambda((0,1])=\lambda(\bigcup_{j=1}^{n}\Big(\tfrac{j-1}n,\tfrac jn]\Big)=\lambda\Big(\bigcup_{j=1}^n(0,\tfrac1n]+\tfrac{j-1}n\Big)=n\,\lambda((0,\tfrac1n]), $$ so $$ \lambda([0,\tfrac1n])=\tfrac1n. $$
Then $$ \lambda((0,\tfrac mn])=\lambda\Big(\bigcup_{j=0}^{m-1}(\tfrac j n,\tfrac{j+1}n]\Big)=\sum_{j=0}^{m-1}\lambda((0,\tfrac 1n]+\tfrac jn)=m\,\lambda([0,\tfrac 1n])=\tfrac mn. $$ That is, $\lambda((0,q])=q$ for any positive $q\in\mathbb Q$. Given $p,q\in\mathbb Q$ with $p<q$, $$ \lambda((p,q))=\lambda((p,q])=\lambda((0,p-q]+q)=p-q. $$ So $\lambda=m$ on the intervals with rational endpoints, that generate $B_{\mathbb R}$. Then$\lambda=m$ on $B_\mathbb R$.