Theorem 5 Let $[a,b]$ be a closed, bounded interval and $1 \le p < \infty$. Suppose that $T$ is a bounded linear functional on $L^p[a,b]$. Then there is a function $g \in L^q[a,b]$, where $q$ is the conjugate of $p$, for which $$T(f) = \int_a^b g \cdot f \mbox{ for all } f \in L^p[a,b]$$
Here is the part I want to focus on:
At this point, the author has already shown that there exists an integral function $g$ over $[a,b]$ for which $T(f) = \int_{a}^{b} g \cdot f$ for all step functions $f$ on $[a,b]$.
My first question is, why can the sequence $\{\varphi_n\}$ be taken as uniformly pointwise bounded on $[a,b]$? I looked at the proof of proposition 10 and discovered uniform boundedness mentioned nowhere. Second, how can we conclude that $$\lim_{n \to \infty} \int_a^b g \cdot \varphi_n = \int_a^b g \cdot f$$ from the Dominated Convergence Theorem if it isn't known that $g \cdot \varphi_n \to g \cdot f$ pointwise a.e. on $[a,b]$? We only have that $\varphi_n \to f$ in $L^p[a,b]$, not pointwise. I imagine that $|g \cdot f |$ is the dominating function, but I'm not sure I see why.
My last question is, why bother doing any of this? If the linear functional $f \mapsto \int_a^b g \cdot f$ agrees with $T$ on the set of step functions, which is does, then they agree on $L^p[a,b]$, since the step functions are dense in $L^p[a,b]$. So what is the point of showing this further fact that they agree on the simple functions?