Let $1 \leq p < \infty$, $(X,\mathcal{M},\mu)$ be a sigma-finite measure space. Let $L$ be a continuous linear form on $L^{p}(X,\mathcal{M},\mu)$. Then, show that $\exists g \in L^{p'}$ such that:
$L(f)=\displaystyle\int fg d\mu \quad \forall f \in L^{p}(X,\mathcal{M},\mu)$
I have shown the result for the underlying space being a finite measure space. If our space is sigma finite, then we can find an increasing sequence of sets $(X_{n})_{n=1}^{\infty}$ such that $X=\cup_{n=1}^{\infty}X_{n}$ for the $X_{n}$s having finite measure.
I was thinking of restricting ourselves to a $X_{n}$ and then finiding $g_{n} \in L^{p'}(X_{n}, \mathcal{M}_{n}, \mu)$, for $g_{n}$ having $L^{p'}$ norm smaller than the operator norm of $L$ and $L(f)=\int fg_{n} d\mu$, $\forall f \in L^{p}(X_{n},\mathcal{M}_{n},\mu)$ as we have shown the result for finite measure spaces.
However, what would be the way to use the above observation to extend the result to $X$? (and finding the appropriate $g \in L^{p}(X,\mathcal{M},\mu)$ such that we have the result). On the other hand, this might seem like a silly question, but how does one use this result to show that the dual space of $L^{p}$ is isomorphic to $L^{p'}$?
Thanks for the help!