MHB Proving Density of $S$ in $L^{p'}(E)$ for $g \in L^p(E)$

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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.
$\int_{E}\left| fg \right| \leq \left| \left| f \right| \right|_{p'} \left| \left| g \right| \right|_p$
 
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Hi joypav,

By density of $S$ in $L^{p'}(E)$, the integral $\int_E fg = 0$ for all $f \in L^{p'}(E)$. Construct $f\in L^{p'}(E)$ such that $fg = \lvert g\rvert^p$ and $\int_E \lvert f\rvert^{p'} \le \int_E \lvert g\rvert^p$. Deduce that $\int_E \lvert g\rvert^p = 0$, i.e., $\|g\|_{L^p(E)} = 0$. Then $g = 0$ in $L^p(E)$.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.

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