Let $g$ be continuous on $[a,b]$. Let $\{f_n\}_{n=1}^\infty$ be a sequence of continuous function, and $\{f_n\}_{n=1}^\infty$ converges to $f$ uniformly on $[a,b]$. Prove that $$\displaystyle \lim_{n\to\infty}\int_a^bf_ng = \int_a^bfg.$$
I'm having some trouble actually proving this result. I had a general outline of what I think needs to happen, but I'm having quite a difficult time filling in the details.
Because $f_n\rightrightarrows f$ (that is, $\{f_n\}_{n=1}^\infty$ converges uniformly to $f$ on $[a,b]$), we can say that given $\epsilon > 0$, there exists $N\in\mathbb N$ such that for all $n\in\mathbb N$ satisfying $n\ge N$, we have $\left|f_n-f\right|<\epsilon/\square.$ So,$$\begin{align}\left|\int_a^bf_ng-\int_a^bf\right| &= \left|\int_a^b(f_n-f)g\right|\\ & < \left|\int_a^b{\epsilon\over \square}|g|\right|\\&={\epsilon\over\square}\left|\int_a^b|g|\right|.\end{align}$$
What happens here though? Do I need to break this into cases? I can't quite make sense of what happens when $\displaystyle \left|\int_a^b|g|\right| =0$. If $\displaystyle \left|\int_a^b|g|\right| > 0$ then we're fine. That's all we need to put in the box. However, if $\displaystyle \left|\int_a^b|g|\right| = 0$, what happens exactly? Was this even the correct approach? I don't feel like I used all the assumptions.
Stick closer to the estimates your textbook gives you. Compare these estimates with your estimates: $$ |\int_a^b f_n g - \int_a^b fg | \leq | \int_a^b (f_n -f)g|\le \int_a^b |f_n-f||g| \leq \int_a^b \epsilon|g| \leq C(a-b) \epsilon $$ Where $C \geq0$ is chosen such that $|g| \leq C$. This is well defined since we have a continuous function on a compact set.
Looking at your estimate, you had to shortly explain why $(f_n-f)g \leq \epsilon |g|$. Also, since your integrand is positive, there is no need for absolute value around the whole expression. More generally, there is the estimate: $$ |\int_a^b f |\leq\int_a^b|f| $$ To answer your other question: If $\int_a^b |g|=0$ and $g$ is continuous, there must be that $g=0$. You can proof this with the help of the $\epsilon -\delta$ defintion of continuity for $|g|$ and monoticity of integrals. Also, I dont see a problem with $g=0$. You dont need to assume $g \neq 0$ at any point. The calculation becomes trivial once you assume $g=0$.