# Measures idempotent with respect to addition and multiplication.

Does there exist a probability finitely additive measure on $\mathbb N$ which is idempotent with respect to addition and multiplication simultaneously?

It is known (due to Hindman) that there is no ultrafilter that is additively and multiplicatively idempotent. But in the closure of the set of additively idempotent ultrafilters there are multiplicatively idempotent ultrafilters.

This question is related to the ones of Justin Moore: Idempotent measures on the free binary system? and Do distinct idempotent measures on finite binary systems have distinct supports?. There is good background in the first references, but I do repeat some definition...

Consider $l_\infty(\mathbb{N})$ with $\mathbb R$ as a base field. A finitely additive probability measure on $\mathbb{N}$ is $\mu\in l_\infty^*(\mathbb{N})$ which is positive ($\mu(f)\geq 0$ if $f\geq 0$) and $\mu(1)=1$. Denote the set of these measures as $PM(\mathbb{N})$. Let $\star=+$ or $\cdot$. We may define $\star$ on $\mu\in l_\infty^*(\mathbb{N})$ as follows: $$\nu\star\mu(f)=\nu_x(\mu_y(f(x\star y))),$$ where $\mu_y(f(x\star y))$ means that we apply $\mu$ with respect to $y$ for $x$-shifts of $f$, the result is a function of $x$. It looks that the only algebraic property of $+,\cdots$ conserved by this extension is associativity. The set $PM(\mathbb{N})$ is closed with respect to $\star$. So, in this notation the question is

Does the exist $\mu\in PM(\mathbb{N})$ such that $\mu+\mu=\mu\cdot\mu=\mu$?

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May be I was to quick to post the question. It has the negative answer: $\mathbb{Z}_2$ has no such measure: it has unique idempotent for $\cdot$: the measure of $1$ is $1$. This is not an idempotent for $+$. The homomorphism $\mathbb{N}\to\mathbb{Z}_2$ solves the problem. So, why the approach of Hindman was so complicated? Probably he thought about ultrafilters only.... – Lev Glebsky Dec 6 '12 at 4:28

I decide to repeat my comments to mark this question as answered. It has the negative answer: If such a measure existed then such a mesure would exist for any homomorphic images of $N$. But ${\mathbb Z}_2$ does not have such a measure.