Real Analysis: Uniform Continuity Here is the question I'm having trouble with:

Suppose that $g$ is uniformly continuous on an interval $I$ and that $f$ is uniformly continuous on an interval $J$ that contains $g(I)$. Prove that the composite function $f  g$ is uniformly continuous on $I$.

First I'm taking the definition of uniform continuity as:

The function $f$ is uniformly continuous on $I$ if for every $ε > 0$, there exists a $δ > 0$ such that:
  $|x − y| < δ$ implies $|f(x) − f(y)| < ε$.

I am able to prove this if the function $f$ and $g$ are defined as uniformly continuous and bounded on the same interval $I$. Here is my proof in that case:
Because $f$ and $g$ are bounded, there is an $M > 0$ such that $|f(x)| < M$ and $|g(x)| < M$ for all $x$ in $I$.
Because $f$ and $g$ are uniformly continuous on $I$, given $ε > 0$, there is a $δ > 0$ such that if $x$ and $y$ are in $I$ and
$|x − y| < δ$, then $|f(x) − f(y)| < ε/2M$ and $|g(x) − g(y)| < ε/2M$. Then:
\begin{align*}
|f(x)g(x) − f(y)g(y)| &= |f(x)g(x) − f(y)g(x) + f(y)g(x) − f(y)g(y)|\\
&\le |f(x)g(x) − f(y)g(x)| + |f(y)g(x) − f(y)g(y)| & \text{(Triangle Inequality)}\\
&\le |g(x)||f(x) − f(y)| + |f(y)||g(x) − g(y)| \\
&\le M|f(x) − f(y)| + M|g(x) − g(y)| & \text{(f and g bounded by M)}\\
& < M (ε/2M) + M(ε/2M) & \text{(Uniform continuity)}\\
&= ε.
\end{align*}
However, I am having trouble expanding this proof to include the case where $f$ is uniformly continuous on an interval $J$ that contains $g(I)$. Please help!
Thanks a lot.
 A: 
Suppose that $g$ is uniformly continuous on an interval $I$ and that $f$ is uniformly continuous on an interval $J$ that contains $g(I)$. Prove that the composite function $f \circ g$ is uniformly continuous on $I$.

First noted that you may type \circ in $\LaTeX$ for the composition operator $\circ$.

Proof.
Since $g$ is uniformly continuous on $I$, by definition, 
Given any $\varepsilon' > 0$, we have for some $\delta'(\epsilon') >0$, $x', x'' \in I $ and $\lvert x' - x'' \rvert < \delta' \Rightarrow \lvert g(x') - g(x'') \rvert < \varepsilon' $.
Now we observe $g(I) \subset J$. It is clear that the restriction of $f$ on $g(I) \subset J$ is also uniform continuous, thus applying the definition, we have: 
Given any $\varepsilon > 0$, $\exists \delta(\epsilon) > 0$, such that $y', y'' \in g(I)$ and $\vert y' - y'' \rvert < \delta \Rightarrow \lvert f(y') - f(y'') \rvert < \epsilon$.
Hence if we choose $\varepsilon' = \delta(\epsilon)$ in the first application, we have $x', x'' \in I$ and $\lvert x' - x'' \rvert < \delta'(\delta(\varepsilon)) \Rightarrow \lvert g(x') - g(x'') \rvert < \delta(\varepsilon)$. Noted $g(x'), g(x'') \in g(I) \subset J$, hence it further implies that $\lvert f(g(x')) - f(g(x'')) \rvert < \varepsilon$.
In other words, we have shown that given any $\varepsilon > 0$, we may find some $\delta > 0$ such that $x' ,x'' \in I$ and $\lvert x' - x'' \rvert < \delta \Rightarrow \lvert f \circ g(x') - f \circ g(x'') \rvert < \varepsilon$. Hence by definition $f \circ g$ is uniform continuous on $I$, as desired.

Noted in this proof we do not need to use the fact that $f$ and $g$ are bounded.
