# Measures on $\mathbb{R}$ that are not translation invariant

I am looking for examples of measures on $\mathbb{R}$ which are not translation invariant.

The only one I could come up so far is the dirac measure.

In particular, I am looking for a measure $\mu$ on the measurable space $(\mathbb{R}, \mathcal{P}(\mathbb{R}))$ which is not invariant under translation but fulfills $\mu((a,a+1]) = 1$ for all $a \in \mathbb{R}$.

Let $\omega\colon\mathbb{R}\to[0,\infty)$ be measurable. Then $$\mu_\omega(A)=\int_A\omega(x)\,dx$$ defines a measure, which in general will not be translation invariant. If moreover $\omega$ is periodic if period $1$ and $\int_0^1\omega(x)\,dx=1$, then $\omega$ will satisfy your requirements.
• The process you describes upfront a measure on all Lebesgue measurable subsets of $\mathbb{R}$, not on $\mathcal{P}(\mathbb{R})$. – Anguepa Jul 4 '17 at 11:30
If it needn't be finite, then $\mu(k)=1$ for $k\in\mathbb{Z}$ (and zero on $\mathbb{R}\setminus\mathbb{Z}$) has got this property. For example $\mu(\{1\})\neq\mu(\{3/2\})$.
• What does $\mu(k) = 1$ mean? The measure is defined on the power set, but I don't understand what the image of a particular subset of $\mathbb{R}$ should be according to your answer. – el_tenedor Jul 31 '15 at 15:25
• @el_tenedor Not on the whole power set, but on some subset of measuarable sets. If we define a (sometimes called a discrete) measure as 1 on every integer and 0 elsewhere, it is defined at least on all Borel subsets of $\mathbb{R}$, because the measure of every open interval is defined.. – Przemysław Scherwentke Jul 31 '15 at 16:10
• Why not on the whole power set? If I define the discrete measure as $$\mu = \sum_{k \in \mathbb{Z}} \delta_k$$ where $\delta_k$ is the dirac measure it should work for every subset. – el_tenedor Jul 31 '15 at 16:21