Proof Verification: Convergence of Riemann integrable function on unbounded domain (Rudin 7.12) Could someone please verify my solution I have for the below problem?

Suppose $g$ and $f_n (n=1,2,3,...)$ are defined on $(0,\infty)$, are Riemann-integrable on $[t,T]$ whenever $0<t<T<\infty$, $\mid f_n \mid \leq g, f_n\rightarrow f$ uniformly on every compact subset of $(0,\infty)$, and
$\int_0^\infty g(x)dx<\infty$
Prove that
$\lim_{n\rightarrow\infty}\int_0^\infty f_n(x)dx=\int_0^\infty f(x)dx$

Proof
From the triangle inequality,
$\mid f_n-f\mid\leq \mid f_n\mid+\mid f\mid\leq g+\mid f\mid$
$\int_0^\infty g(x)dx<\infty$ implies that $g$ is a bounded function.
Therefore there exists a $M$ such that $g<M$ for all $x$.
Next, we create a sequence $\{\epsilon_n\}_{n=1}^\infty$ such that 
$\epsilon_n\in(\mid f_n-f\mid,M+\mid f\mid)\neq \varnothing$.
This sequence will converge to $0$ as $n\rightarrow 0$.
Next,
$\int_0^\infty f_n-\epsilon_n dx\leq\int_{-0}^\infty f dx\leq\int_{0}^{-\infty} f dx\leq \int_0^\infty f_n+\epsilon_n dx$.
The two inner integrals denotes the upper/lower Riemann integral.
Since, $2\int_0^\infty \epsilon_n \rightarrow 0$, the Riemann integral on $(0,\infty)$ of $f$ exists.
Q.E.D
 A: The problem is that $g$ may not be bounded. For example, $\int^{\infty}_0x^{-1/2}e^{-x}dx$ exists. 
Here is a sketch of a possible proof of Rudin's exercise:
$\left|\int^b_af_n-f\right|\le 2\int^b_ag \tag1$
There are $0<c<a<b<e<d$ such that 
$\left|\int^d_cg-\int^{\infty}_0g\right|<\epsilon/2 \tag2$
from which it follows by the triangle inequality that 
$\int^e_dg<\epsilon \tag3$
So, we have now since $|f_n|\le g,$
$\left|\int^d_cf_n-\int^e_cf_n\right|<\epsilon \tag4$
from which we conclude that the improper integrals $\int^{\infty}_cf_n$ exist for all integers $n.$ And the same argument applies to $f$ since $|f|\le g$.
So, we can choose $d$ such that 
$\left|\int^d_cf_n-\int^{\infty}_cf_n\right|<\epsilon\ \text{and}\ \left|\int^d_cf-\int^{\infty}_cf \right|<\epsilon \tag5$.
To finish, use an $\epsilon/3$ argument on  
$\left|\int^{\infty}_0f_n-\int^{\infty}_0f \right|\le \left|\int^{\infty}_0f_n-\int^{d}_cf_n \right|+\left|\int^{d}_cf_n-\int^{d}_cf \right|+\left|\int^{\infty}_0f-\int^{d}_cf \right| \tag6$
