Question from Folland theorem 6.15's proof

Question

I am confused with the last sentence, how to follow proposition 6.7 to get $\phi(f)=\int fg$ for all $f\in L^p$.

The proposition 6.7 is in the below:

Answer 1

Let the linear operator $T:S\rightarrow{\bf{C}}$, $T(f)=\displaystyle\int fg$ be defined on the dense proper subspace $S$ of $L^{p}$, where $S$ is the set of all simple functions of the form as in Proposition 6.7.

Now $T$ is bounded because of the inequality that $\left|\displaystyle\int fg\right|\leq\|f\|_{L^{p}}\|g\|_{L^{q}}$. So $T$ has a unique bounded extension $U$ to all of $L^{p}$. Now $U=\phi$ on $S$, by uniqueness, we must have $U=\phi$.

Now the linear operator $V:L^{p}\rightarrow{\bf{C}}$ defined by $V(f)=\displaystyle\int fg$ is bounded also by the inequality, and its restriction to the subspace $S$ is exactly $T$, by uniqueness, we must have $V=U$. So we conclude that $\phi=U=V$.