Bounds for the integral of a function with respect to an invariant measure 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
 A: If you look at the context of $T$ a continuous map from a topological space $X$ to itself and $f$ a continuous function from $X$ to $\mathbb R$, the natural analagous version of the question is looking for invariant ($\sigma$-additive) probability measures on $X$ maximizing or minimizing $\int f\,d\mu$. 
This area of study has received some attention and is called ergodic optimization. See here for a survey. A central conjecture in this area concerns the case $X=\lbrace 0,1\rbrace ^{\mathbb Z^+}$; $T$ is the shift map on $X$ and $f$ a Lipschitz function. It is conjectured that for a residual set of Lipschitz functions, the maximizing measure is supported on a periodic orbit.
A: 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 $\overline\alpha(f)=L(f)=\int f\,d\mu$.
