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Theorem. If $a < b < c$ and the sequence $f_1,\,f_2,\,f_3,\,\ldots$ 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 interval $[a,\,c]$ .
Proof. We have the limit functions $\displaystyle f_{ab} := \lim_{n\to\infty}f_n$ on $[a,\,b]$ and $\displaystyle f_{bc} := \lim_{n\to\infty}f_n$ . It follows that $$f_{ab}(b) = \lim_{n\to\infty}f_n(b) = f_{bc}(b).$$ Define the new function
Choose an arbitrary positive number $\varepsilon$ . The supposed uniform convergences on the intervals $[a,\,b]$ and $[b,\,c]$ imply the existence of the numbers $n_1(\varepsilon)$ and $n_2(\varepsilon)$ such that $$|f_n(x)-f(x)| < \varepsilon\;\;\forall x\,\in [a,\,b],\quad\mbox{when}\; n > n_1(\varepsilon)$$ and $$|f_n(x)-f(x)| < \varepsilon\;\;\forall x\,\in [b,\,c],\quad\mbox{when}\; n > n_2(\varepsilon).$$
If one takes $n > \max\{n_1(\varepsilon),\,n_2(\varepsilon)\}$ , then one has simultaneously on both intervals $[a,\,b]$ and $[b,\,c]$ , i.e. on the whole greater interval $[a,\,c]$ , the condition $$|f_n(x)-f(x)| < \varepsilon.$$
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![$\displaystyle f(x) := \begin{cases}f_{ab}(x) \quad \forall x\,\in [a,\,b],\\ f_{bc}(x) \quad \forall x\,\in [b,\,c]. \end{cases}$](http://images.planetmath.org:8080/cache/objects/9834/js/img1.png "$\displaystyle f(x) := \begin{cases}f_{ab}(x) \quad \forall x\,\in [a,\,b],\\ f_{bc}(x) \quad \forall x\,\in [b,\,c]. \end{cases}$")