Let $f:\mathbb N\rightarrow\mathbb R$ be bounded. Let $\mu$ be a translation invariant finitely additive probability measure on $\mathbb N$. 


**Question:** Are there any lower and upper bounds for $\int fd\mu$?


If $f$ is a characteristic function (or more generally if the image of $f$ is finite), one gets the lower and upper Banach density.


More general results? References? 


Thanks in advance,
Valerio



Bounds for the integral of a function with respect to an invariant measure
Define
$$
\overline\alpha(f)=\limsup\_{n-m\to\infty} \frac{1}{n-m}(f(m)+\ldots+f(n-1))
$$
and
$$
\underline\alpha(f)=\liminf\_{n-m\to\infty} \frac{1}{n-m}(f(m)+\ldots+f(n-1)).
$$


Claim $\max\int f\,d\mu=\overline\alpha(f)$ where the maximum is taken over translation-invariant finitely additive probability measures and $\min\int f\,d\mu=\underline\alpha(f)$.


Clearly it's sufficient to prove it for $\overline\alpha(f)$. Let $M=\overline\alpha(f)$ and let $\epsilon > 0$. By definition of $\limsup$, there exists an $N$ such that for every $n$, $(1/N)(f(n)+\ldots+f(n+N-1)) < M+\epsilon$. Hence we see
$$
\int (1/N)(f(n)+\ldots+f(n+N-1))\,d\mu(n) < M+\epsilon.
$$
By translation-invariance, this gives $\int f(n)\,d\mu(n)\le M$. 


Conversely, let $n\_i-m\_i\to\infty$ be such that $(f(m\_i)+\ldots+f(n\_i-1))/(n\_i-m\_i)\to\overline\alpha(f)$. Define a family of linear functionals $A\_i\colon l^\infty\to\mathbb R$ by $A\_i(g)=(g(m\_i)+\ldots+g(n\_i-1))/(n\_i-m\_i)$. Notice that 
$ \vert A\_i(g) \vert \le \parallel g \parallel\_{\infty}$. 
Let $Y=\lbrace g \in l^{\infty} \colon \lim\_{i\to\infty} A\_i(g) \text{ exists} \rbrace$. 


This is clearly a subspace. Define a linear operator $L$ on $Y$ by $L(g)=\lim\_{i\to\infty} A\_i(g)$. By the above, this has norm at most 1. Applying $L$ to a constant function, we see that $L$ has norm 1. By Hahn-Banach, notice that $L$ can be extended to a norm 1 operator on all of $l^{\infty}$. We claim that $L$ is positive in the sense that $L(h)\ge 0$ when $h\in l^\infty$ is a non-negative sequence. Supposing that $L(h) < 0$ for some non-negative $l^{\infty}$ sequence $h$ (which we can assume to be of norm 1 without loss of generality), then the sequence $k(n)=1-h(n)$ is another $l^\infty$ sequence and $\parallel k \parallel\_{\infty} \le 1$. However we have that $L(k)=L(1)-L(h) > 1$. This contradicts the fact that $\parallel L\parallel\_{\infty}=1$. 


We need to show that $L$ is invariant, that is $L(S(h))=L(h)$ for all $h$ in $l^\infty$ where $S$ is the right shift operator. To see this, notice that $A\_i((S(h)-h))\to 0$ for all $h\in l^\infty$ so that $S(h)-h\in Y$ and $L(S(h))=L(h)$.


We define a measure $\mu$ by $\mu(A)=L(1\_A)$. This immediately defines a finitely-additive probability measure. By the invariance of $L$ it's shift-invariant. By definition of $L$, we have $\overline\alpha(f)=L(f)=\int f\,d\mu$.


