An analysis problem: limit for every $h$ implies the existence of a 'uniform' limit Suppose that $f : \mathbb{R}^{+} \to \mathbb{R}$ is a continuous, and for any $h>0$, $\lim\limits_{x \to \infty } (f(x+h)-f(x)) = 0$. Show that for any given interval $[a,b], a>0$ and $\epsilon > 0$, there exists $M>0$ such that for any $x > M$ and $h \in [a,b]$ we have $\left| f(x+h) - f(x) \right| < \epsilon$.
I have tried this problem but find that the main obstacle is that the condition is hard to use, and it's almost impossible to pass from a 'pointwise' condition to a 'uniform' conclusion. Someone tells me that the proof requires measure theory, but I cannot see the relation. Can anyone give some ideas ?

Here is a sketch of a proof using measure theory. It is somewhat intricate and uses both the Egoroff and Steinhaus theorems.
Consider the montonically decreasing sequence of measurable functions 
$$\phi_n(h) = \sup_{x \geqslant n} |f(x+h) - f(x)|,$$
where $\phi_n(h) \downarrow 0$ as $n \to \infty$.
By Egorroff's theorem there exists a measurable set $E \subset [-1,1]$ with $m(E) > 1$ such that $\phi_n(h)$ converges uniformly on $E$. The set $F = E \, \cap \, \{-x: x \in E\}$has positive measure and by the Steinhaus theorem it can be shown that there exists an interval $[-\delta,\delta] \subset F$ containing $0$ such that any point $h \in [-\delta , \delta]$ is given by $h = s-t$ where $s,t \in F$.
For any $\epsilon > 0$ and all $h = s-t \in [0,\delta]$  there exists $N \in \mathbb{N}$ such that  for all $n > N$ we have
$$\phi_n(h) \leqslant \phi_n(s) + \phi_n(-t) < 2\epsilon,$$
whence, $\phi_n(h) \to 0$ uniformly for $h \in [0,\delta]$.  This can be extended to prove uniform convergence on any compact interval $[a,b]$.
