Suppose $f_n,f,g$ are regulated functions on $[0,R]$ for every $R>0$ and satisfy:
(i)$f_n \to f$ uniformly on $[0,R]$, (ii) $|f_n(x)| \leq g(x)$ on $[0,\infty)$, and (iii) $\int^{\infty}_0g$ is convergent
Show that the improper integral $\int^{\infty}_0f$ exists and $\int^{\infty}_0f_n\to\int^{\infty}_0 $ as $n\to\infty$
A regulated function $f$ on $[a,b]$ is a function for which there exists a step function on $[a,b]$ whiih converges uniformly to $f$
So, my attempt is:
Using the fact that $\int^{\infty}_0g$ converges, $\int^{\infty}_0f_n$ converges and $\int^{\infty}f$ converges $\Rightarrow \int^{\infty}_0f$ exists
Now, $\int^{\infty}_0g $ is convergent $\rightarrow \int^{\infty}_Rg\leq\epsilon$ for some value of $R$ being big enoguh and $\forall \epsilon >0$
Now, proving that $\int^{\infty}f_n\to\int^{\infty}_0f$
$$|\int^{\infty}_0f-\int^{\infty}_0f_n|=|\int^R_0f+\int^{\infty}_Rf-(\int^R_of_n+\int^{\infty}_Rf_n)|\leq |\int^R_0(f-f_n)|+|\int^{\infty}_Rf|+|\int^{\infty}_Rf_n|\leq|\int^R_0(f-f_n)+\int^{\infty}_Rg+\int^{\infty}_Rg\leq3\epsilon$$
Note that $|\int^R_0(f-f_n)|\leq\epsilon \forall \epsilon$ as $f_n\to f$ uniformly
Also note that $|\int^{\infty}_Rf|,|\int^{\infty}_R|\leq\int^{\infty}_Rg$ as $\int^{\infty}_R $ exists $|f_n|,|f|\leq g$
Is my proof correct?