$g:I\rightarrow \mathbb{R}$ for $I\in [0,\infty]$ continuous $g:I\rightarrow \mathbb R$  for $I\in [0,\infty)$ continuous function, for $\epsilon > 0$ $\exists$ $M_\epsilon >0\; \mid \; a>M_{\epsilon} \implies g(a)<\epsilon$ show uniform continuity on $I$
My attempt:  I am still learning.  I need help understanding what this problem is asking me (I don't know what is meant by $M_{\epsilon}$), and what theorems/proofs are available to solve it.
 A: For every $\epsilon>0$, we are given a constant $M(\epsilon)$ such that $g(x)<\epsilon$ for all $x>M(\epsilon)$.
Uniform continuity means that for every $\epsilon$, there is a $\delta$ such that $|g(x)-g(y)|<\epsilon$ whenever $|x-y|<\delta$ and $x,y\in I$.
Proof:
Fix any $\epsilon>0$. Let $J=[0,M(\epsilon/2)+1]$. Observe that $J$ is a compact set, by Heine-Borel. Since $g$ is continuous and $J$ is compact, it follows by a well known theorem that $g$ is uniformly continuous on $J$. Hence there is a $\delta$ for which $|g(x)-g(y)|<\epsilon/2$ whenever $|x-y|<\delta$ and $x,y\in J$.
Now let $x,y\in I$ be any points such that $|x-y|<\min\{\delta,1\}$. We claim that $|g(x)-g(y)|<\epsilon$. Indeed, if both $x,y\in J$ the result is already shown, so we may suppose $x\not\in J$, i.e. $x\geq M(\epsilon/2)+1$. Since $|x-y|<1$, it follows by the triangle inequality that $y\geq M(\epsilon/2)$. Thus by the definition of $M(\epsilon/2)$, it follows that $g(y)<\epsilon/2$ and the same bound holds for $g(x)$. Once more by the triangle inequality, we have $|g(x)-g(y)|\leq |g(x)|+|g(y)|<\epsilon$.
Therefore we have shown that in all cases, $|x-y|<\min\{\delta,1\}$ implies that $|g(x)-g(y)|<\epsilon$. Since $\epsilon>0$ was arbitrary, this completes the proof.
