A question regarding the proof of Riesz Representation Theorem for the dual of Lp. The theorem is stated below.

Pasted following is the proof. Can anyone explain why $g \rightarrow S(g)-\int_X fgd\mu \  for\ all\ g\in L^p$ is continuous?
And right above equation 15, why is $|g_n - g| <= |g|^p$?
Where in the proof is it showing that the mapping is isometric and isomorphic?


 A: 
Why is
  $$g\mapsto S(g)-\int\limits_{X}{fg\text{ d}\mu}\text{ for all }g\in L^p$$
  continuous?

By assumption, $S$ is a bounded linear functional on $L^p$. Since $f\in L^q$, as established in the previous line, by Holder's inequality we have
$$\left|\int\limits_{X}{fg\text{ d}\mu}\right|\le\|f\|_{L^q}\|g\|_{L^p}.$$
Hence, $g\mapsto\int\limits_{X}{fg\text{ d}\mu}$ is a bounded linear functional on $L^p$. Thus, $g\mapsto S(g) - \int\limits_{X}{fg\text{ d}\mu}$ is also a bounded linear functional on $L^p$, and hence continuous.

Right above equation (15), why is $|g_n-g|^p\le |g|^p$?

By construction, $g_n = g$ on $X_n$ and $g_n = 0$ on $X\backslash X_n$. Thus, $g_n-g = 0$ on $X_n$ and $g_n-g = -g$ on $X\backslash X_n$, so $|g_n-g|^p\le |g|^p$ on $X$, as $|g_n-g|^p$ at any point is either $0$ or $|g|^p$.

Where in the proof is it showing that the mapping is isometric and isomorphic?

Presumably, equation (8) says something like the following: for $f\in L^q(X,\mu)$, define $T_f\in (L^p(X,\mu))^*$ by
$$T_f(g) = \int\limits_{X}{fg\text{ d}\mu}\text{ for all }g\in L^p(X,\mu).$$
Let $T:L^q(X,\mu)\rightarrow(L^p(X,\mu))^*$ by $T(f) = T_f$. Clearly, $T$ is linear. Now, $T$ is surjective, since the above proof shows that for every $S\in (L^p(X,\mu))^*$, there exists $f\in L^q(X,\mu)$ such that $T_f = S$. To show that $T$ is an isometry from $L^q(X,\mu)$ to $(L^p(X,\mu))^*$, we must show that $\|T_f\|_{(L^p)^*} = \|f\|_{L^q}$ for every $f\in L^q$. Now by Holder's inequality
$$\|T_f\|_{(L^p)^*} = \sup\limits_{\|g\|_p=1}{\langle T_f, g\rangle} = \sup\limits_{\|g\|_p=1}{\int\limits_{X}{fg\text{ d}\mu}}\le\sup\limits_{\|g\|_p=1}{\|f\|_{L^q}\|g\|_{L^p}} = \|f\|_{L^q}$$
so $\|T_f\|_{(L^p)^*}\le \|f\|_{L^q}$ for all $f\in L^q$. However, the above proof also showed that if $S\in (L^p)^*$ and $S = T_f$, then $\|f\|_{L^q}\le\|S\|_{(L^p)^*}$. In particular, $\|f\|_{L^q} \le \|T_f\|_{(L^p)^*}$ for all $f\in L^q$. This implies $\|T_f\|_{(L^p)^*} = \|f\|_{L^q}$ for all $f\in L^q$, and so $T$ is indeed an isometry from $L^q(X,\mu)$ to $(L^p(X,\mu))^*$. I leave it to you to show why $T$ is an isomorphism, if it is not already clear by now.
