Probability measure satisfying translation-invariance property I have a probability space $(\mathbb {Z},2^{\mathbb {Z}},\mathbb {P})$. I'm thinking about a counter example if the probability measure satisfies the translation-invariance property, which means $\mathbb {P}(E+n)=\mathbb {P}(n)$ for every event $E\in 2^{\mathbb {Z}}$ and $n\in \mathbb {Z}$, and $E+n$ is obtained by adding $n$ to every element of $E$. I wonder which property of the probability measure it violates if such a $\mathbb {P}$ exists. I'm thinking about letting $E=2\mathbb {Z}$ and $n=1$, but I'm not sure if I'm on the correct track.
 A: With your example $E=2\mathbb Z$, $E$ and $E+1$ form a partition of $\mathbb Z$ hence these sets would have probability $1/2$ and there would be nothing wrong with it a priori.
Instead, let $E=\{0\}$; if the translation invariance property is satisfied, then
$\mathbb P(\{n\})=\mathbb P(\{0\})$ for each $n$ and by $\sigma$-additivity, the series $1=\sum_{n\in\mathbb Z}\mathbb P(\{n\})=\sum_{n\in\mathbb Z}\mathbb P(\{0\})$. This cannot happen if $\mathbb P(\{0\})=0$; if $\mathbb P(\{0\})>0$ the series $\sum_{n\in\mathbb Z}\mathbb P(\{0\})$ diverges hence we also get a contradiction.
In other words, translation invariance in this setting is not compatible with $\sigma$-additivity.
Note that a similar argument shows that there is no translation invariant probability measure on $\mathbb R$ endowed with the Borel $\sigma$-algebra: let $E=[0,1)$; if $\mu$ were such a measure, then $1\geqslant\mu([0,N))\sum_{n=0}^{N-1}\mu(n+[0,1))=N\mu([0,1))$ hence $\mu([0,1))=0$ but then again by translation invariance each interval would have measure zero.
