Here is the proof that an orientation preserving homeomorphism $f$ of $\mathbb T$ has a well-defined rotation number. Let $F:{\mathbb R}\rightarrow\mathbb R$ be its lift. It is an increasing function verufying $F(x+\ell)=F(x)+\ell$ for every integer $\ell$. Let $x\in\mathbb R$ be given and $u_n=F^{(n)}(x)$. We have to prove that $\frac1nu_n$ has a finite limit. To do so, fix $n$ and define $N$ so that $u_n\in[x+N,x+N+1)$. Then $$u_{n+m}=F^{(m)}(u_n)\in[F^{(m)}(x+N),F^{(m)}(x+N+1))=[F^{(m)}(x)+N,F^{(m)}(x)+N+1).$$ This gives $u_{n+m}\in[u_m+N,u_m+N+1)$. Consequently, we obtain $$u_m+u_n-1-x\le u_{n+m}\le u_m+u_n+1-x.$$ Applying Fekete's Lemma to $v_n=u_n+1-x$, we see that $\frac1nv_n$, hence $\frac1nu_n$, has a limit $\rho<+\infty$. Applying it to $w_n=u_n-1-x$, we see that this limit is finite.

Finally, the limit does depend upon the starting point $x$, because if $x\le y\le x+1$, then $F^{(m)}(x)\le F^{(m)}(y)\le F^{(m)}(x)+1$. An other use of the monotonicity shows that the limit is the same as $n\rightarrow-\infty$.

Here is the proof that an orientation preserving homeomorphism $f$ of $\mathbb T$ has a well-defined rotation number. Let $F:{\mathbb R}\rightarrow\mathbb R$ be its lift. It is an increasing function verifying $F(x+\ell)=F(x)+\ell$ for every integer $\ell$. Let $x\in\mathbb R$ be given and $u_n=F^{(n)}(x)$. We have to prove that $\frac1nu_n$ has a finite limit. To do so, fix $n$ and define $N$ so that $u_n\in[x+N,x+N+1)$. Then $$u_{n+m}=F^{(m)}(u_n)\in[F^{(m)}(x+N),F^{(m)}(x+N+1))=[F^{(m)}(x)+N,F^{(m)}(x)+N+1).$$ This gives $u_{n+m}\in[u_m+N,u_m+N+1)$. Consequently, we obtain $$u_m+u_n-1-x\le u_{n+m}\le u_m+u_n+1-x.$$ Applying Fekete's Lemma to $v_n=u_n+1-x$, we see that $\frac1nv_n$, hence $\frac1nu_n$, has a limit $\rho<+\infty$. Applying it to $w_n=u_n-1-x$, we see that this limit is finite.

Finally, the limit does not depend upon the starting point $x$, because if $x\le y\le x+1$, then $F^{(m)}(x)\le F^{(m)}(y)\le F^{(m)}(x)+1$. An other use of the monotonicity shows that the limit is the same as $n\rightarrow-\infty$.