Let $E$ be a measurable set, $1 \le p \le \infty$, $q$ the conjugate of $p$, and $S$ a dense subset of $L^q(E)$. If $g \in L^p(E)$ and $\int_E fg = 0$ for all $f \in S$, then $g=0$.
I was able to prove the $p < \infty$ case, but I am having trouble with the $p=\infty$. Here is my proof for $p < \infty$. First, recall the following (I already proved the following):
For $1 \le p \le \infty$, $q$ the conjugate of $p$, and $g \in L^p(E)$, show that $g=0$ if and only if $$\int_E fg = 0 \mbox{ for all } f \in L^q(E)$$
Thus, given $f \in L^q(E)$, there exists a sequence $\{f_n\} \subseteq S$, since $S$ is dense, for which $\|f_n - f\|_q \to 0$ as $n \to \infty$. Then $$\int_E fg = \int_E (f - f_n + f_n) g = \int_E (f-f_n)g + \int_E f_n g = \int_E (f-f_n)g \le \|f_n - f\|_q \|g\|_p$$
which is true by Holder's inequality. Letting $n \to \infty$, we obtain $\int_E fg = 0$. Since $f$ was arbitrary, $g=0$.
How does this sound? As I said above, I am having trouble with the $p = \infty$ case and could use a hint.
Let $g_{n}=\text{sgn}(f)\chi_{\{|x|\leq n, |f|\leq n\}}$, where $\text{sgn}(f)(x)=\overline{f(x)}/|f(x)|$ for $f(x)\ne 0$ and $\text{sgn}(f)(x)=0$ for otherwise, then $g_{n}\in L^{1}$ and we can find a $\varphi_{n}\in S$ such that $\|g_{n}-\varphi_{n}\|_{L^{1}}<1/n$, then we have \begin{align*} \left|\int_{E}fg_{n}\right|&\leq\left|\int_{E}f(g_{n}-\varphi_{n})\right|+\left|\int_{E}f\varphi_{n}\right|\\ &\leq\dfrac{1}{n}\|f\|_{L^{\infty}}. \end{align*} But $0\leq fg_{n}(x)\uparrow|f(x)|$ a.e. $x$ as $n\rightarrow\infty$ since $f\in L^{\infty}$ implies that $|f(x)|<\infty$ a.e., so we have by Monotone Convergence Theorem that \begin{align*} \int_{E}|f|=\lim_{n\rightarrow\infty}\int_{E}fg_{n}=0. \end{align*}