Let $\mu$ be a measure on the borel $\sigma$-algebra of $\mathbb{R}$ such that $\mu([a,b]) = b-a$ for any closed interval. I want to show that $\mu$ is translation invariant i.e for every borel set $E$ and for every $x \in \mathbb{R}$ $\mu(E) = \mu(E+x)$.
I'm trying to prove that by showing that the set $\mathcal{A}=\{E|\forall x\in\mathbb{R}:\mu(E)=\mu(x+E)\}$ $\,$ is a $\sigma$-algebra but because he is closed under countable disjoint union it's enough to prove he is an algebra.
I saw the question over here A Borel measure s.t. $\mu([a,b])=b-a$ is translation invariant but I can't use the $ π–λ$ theorem.
The $\emptyset\in\mathcal{A}$ $\,$ because $\emptyset = \emptyset + x\ \forall x\in \mathbb{R}$
We need to show that $\mathcal{A}$ is closed under complements and union of 2 sets thats where I got stuck. What I tried: Let $E\in\mathcal{A}$ $\ $ then $\mu(E)=\mu(E+x) \ \forall x \in\mathbb{R}\ $and $E \cup E^c = \mathbb{R}=(E+x)\cup (E+x)^c=(E+x)\cup(E^c+x)\ $ so if $\mu(E)<\infty\ $we have $\infty = \mu(\mathbb{R})=\mu((E+x)\cup(E^c+x))=\mu(E+x)+\mu(E^c+x)\ $ so we get $\mu(E^c+x)=\infty\ $ similarly we can get $\mu(E^c) = \infty\ $ so we have $\mu(E^c)=\mu(E^c+x)\ $ but I'm stuck on the cases $\mu(E)=\mu(E^c) = \infty\ $ and $\mu(E)=\infty \, \mu(E^c)<\infty$.
For the union I tried to take two sets $A\ $ and $B\ $ and write them as a disjoint union and use the propeties of a measure but I got nowhere.Any ideas how to proceed?