I know this question has been asked before, but I wanted to try a different proof and get help tying up a piece of the proof that wasn't clear to me in that answer for $p < \infty$ (basically it seemed like the reasoning of that OP only showed $\int_{E} fg \leq 0$ where we need it to be equal to 0). The problem as it was given to me expects me to apply the Riesz Representation Theorem, so here is my attempt:
Let $E$ be measurable, $1 \leq p < \infty$, $q$ the conjugate of $p$, and $S$ dense in $L^{q}(E)$. We want to show that if $g \in L^{p}(E)$ and $\int_{E} fg = 0$ for all $f \in S$, then $g = 0$.
Now, by Riesz Rep, we know that in particular for the Id operator, there exists unique $f \in L^{q}(E)$ such that $Id(g) = g = \int_{E} gf$ for all $g \in L^{p}(E)$.
Since $S$ is dense in $L^{q}(E)$, there exists a sequence $\{s_{n}\} \in S$ such that the limit as n goes to $\infty$ of $||f- s_{n}||_{q}$ is 0. Thus:
$g = Id(g) = \int_{E} gf = \int_{E} g[f - s_{n} + s_{n}] = \int_{E}g[f - s_{n}] + \int_{E} gs_{n} \leq ||g||_{p}||f - s_{n}||_{q}$, which goes to 0.
My hangup: All this proves is that $g \leq 0$. How do we prove $g \geq 0$?