# Uniform convergence on every bounded closed intervals implies uniform convergence on $\Bbb R$

When I read my lecture notes, I found that the outline of the proof for uniform convergence of cosine function$$\cos x =1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots$$on $\Bbb R$ is as the following:

On $[-M,M]\ (M>0)$, the cosine function is uniformly convergent on $[-M,M]$ by M-Test. Since $M$ is arbitrary, it follows that it is uniformly convergent on $\Bbb R$.

But I am doubting that although I can adjust $N$ such that $||f_n-f||\lt ε$ for $n\ge N$ on $[-M,M]$, where $\{f_n\}$ is the sequence of functions and $f_n$ converges uniformly to $f$, I need to choose a bigger $N$ for lager interval, like I get $N$ for $[-M,M]$ and I may need to take $N+1$ or even lager for $[-M-1,M+1]$, so it seems that I cannot find a $N$ such that $||f_n-f||\lt ε$ for $n\ge N$ on $\Bbb R$.

• It's not true. Take $f_n=\chi_{[-n,n]}$. This converges to the constant function $1$ uniformly on any closed, bounded interval. Commented Apr 30, 2016 at 10:28
• @DavidMitra Can I add other assumption to make "Uniform Convergence on every bounded closed intervals implies Uniform Convergence on $\Bbb R$" true? Commented Apr 30, 2016 at 10:36