S is a dense subset of $L^{p'}$, $\int_{E}fg = 0$ for all $f \in S$, then $g= 0$ Problem:
$E$ is a measurable set and $1 \leq p < \infty$. Let $p′$ be the conjugate of $p$, and $S$ is a dense subset of $L^{p′}(E)$.  Show that if $g \in L^p(E)$ and $\int_{E}fg = 0$ for all $f \in S$, then $g= 0$.
Definition of Density:
$S$ is dense in $L^{p'}(E)$ if $\forall h \in L^{p'}(E), \forall \epsilon > 0, \exists f \in S$ s.t. $\left| \left| f-h \right| \right|_{p'} < \epsilon$
or equivalently
$\exists (f_n)$ in $S$ s.t. $\lim_{{n}\to{\infty}}f_n=h$ a.e. on $E$.
Idea?:
As p and p' are conjugates, I was thinking to use Holder's Inequality.
$\int_{E}\left| fg \right| \leq \left| \left| f \right| \right|_{p'} \left| \left| g \right| \right|_p$

Notice that
$$
X=\{f \in L^{p'}\ |\ \int_E fg=0\}
$$ is a closed subspace as a kernel of the continuous linear functional $L:f\mapsto \int_E fg$. By the assumption that $S\subset X$ and $\overline{S}=L^{p'}$, we have that$$
L^{p'} =\overline{S} \subset X,
$$ that is, $X=L^{p'}$. This gives $L\equiv 0$ and by Riesz representation theorem, we have
$$
\|g\|_{L^{p}}=\|L\|=0,
$$ i.e. $g =0$ in $L^{p}$.
