uniform convergence on union interval

Theorem. If    and the sequence  $f_1,f_2,f_3,\mathrm{\dots }$  of real functions converges uniformly both on the interval  $[a,b]$  and on the interval  $[b,c]$,  then the function sequence converges uniformly also on the union (http://planetmath.org/Union) interval  $[a,c]$.

Proof. We have the limit functions $f_{ab}:=\underset{n\to \mathrm{\infty }}{lim}{f}_n$  on  $[a,b]$  and $f_{bc}:=\underset{n\to \mathrm{\infty }}{lim}{f}_n$. It follows that

$$f_{ab}(b)=\underset{n\to \mathrm{\infty }}{lim}{f}_n(b)=f_{bc}(b).$$

Define the new function

$f(x):=\{\begin{array}{cc}{f}_{ab}(x)\hspace{1em}\forall x\in [a,b],\hfill & \\ f_{bc}(x)\hspace{1em}\forall x\in [b,c].\hfill & \end{array}$

Choose an arbitrary positive number $\epsilon$. The supposed uniform convergences on the intervals  $[a,b]$  and  $[b,c]$  imply the existence of the numbers $n_1(\epsilon )$ and $n_2(\epsilon )$ such that

and

If one takes  $n>\max \{n_1(\epsilon ),n_2(\epsilon )\}$,  then one has simultaneously on both intervals  $[a,b]$  and  $[b,c]$,  i.e. on the whole greater interval  $[a,c]$,  the condition