# Why this function is uniformly continuous at $(a,b)$

I saw a theorem that say:
"Let $f:\left(a.b\right)\rightarrow \mathbb{R}$ be continuous function.
So, $f$ is uniformly continuous in $(a,b)\Leftrightarrow$ $\lim\limits_{x\to a^+}f(x)$ and $\lim\limits_{x\to b^-}f(x)$ exist".

I have problem with going from left to right side.

I saw a proof here Show f is uniformly continuous on $(a,b)$ if it is continuous and $\lim\limits_{x\to a^+}f(x)$ and $\lim\limits_{x\to b^-}f(x)$ exist.
But unfortunately the user who answered there is no longer in this website (it's been 2 years from this question) so its Not "duplicating", its just find another answer or gudiance.I'll copy his answer:

Extend the function $f$ to the function $g:[a,b]\to\mathbb R$ with $g(a)=\lim_{x\to a^+} f(x)$ and $g(b)=\lim_{x\to b^-}f(x)$. Then $g$ is continuous on $[a,b]$. It follows that $g$ is uniformly continuous on $[a,b]$ so that $f$ is uniformly continuous on $(a,b)$.

Why does "$g$ is uniformly continuous on $[a,b]$ so that $f$ is uniformly continuous on $(a,b)$."?

The definitions of uniform continuity for $g$ and for $f$ are the same except that for $f$ one needs to check it for less points ($a,b$ don't need to be checked). Therefore if $g$ is uniformly continuous so is $f$.
If $f$ is uniformly continuous then for all $\epsilon>0$ there is $\delta$ such that if $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon$. Let $x_n$ be a sequence of points tending to $b$. Then there is $N$ such that for all $n,m>N$ we have that $|x_n-x_m|<\delta$. Therefore $|f(x_n)-f(x_m)|<\epsilon$. This means that $f(x_n)$ converges. Since this is true for all sequences $x_n$ that tend to $b$, then $\lim_{x\to b}f(x)$ exists.
A similar argument shows that $\lim_{x\to a}f(x)$ exists.
Since $g$ is uniformly continuous on $[a,b]$, given $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x,y\in [a,b]$, $|x - y| < \delta$ implies $|g(x) - g(y)| < \epsilon$. In particular, for all $x,y\in (a,b)$, $|x - y| < \delta$ implies $|f(x) - f(y)| = |g(x) - g(y)| < \epsilon$. Since $\epsilon$ was arbitrary, $f$ is uniformly continuous on $(a,b)$.
Also, to prove the forward direction, suppose $f$ is uniformly continuous on $(a,b)$. Let $\{a_n\}$ be a sequence in $(a,b)$ with with $\lim a_n = a$. Since $\{a_n\}$ is Cauchy sequence and $f$ is uniformly continuous, $f(a_n)$ is a Cauchy sequence. So $\{f(a_n)\}$ is convergent in $\Bbb R$. Since $\{a_n\}$ was arbitrary, it follows that $\lim_{x\to a^+} f(x)$ exists. By a similar argument, $\lim_{x\to b^{-}} f(x)$ exists.