Showing that $\int fg\le \int g$ implies $f=0$ a.e. Take $0<p<1$. If $f$ is locally integrable over on $\mathbb{R}$ and $$\Bigg\vert \int fg\Bigg\vert\le \Vert g\Vert_p\tag 1$$ for every $g$ continuous on a set of compact support, then $f=0$ a.e.
By Holder $\Bigg\vert \int fg\Bigg\vert\le \Vert f\Vert_q\Vert g\vert_p$. This seems weaker than $(1).$ Plugging in $\chi _E=g$ for any compact set $E$ containing the domain of $f$ we have $\vert \int f\vert=\vert \int fg\vert\le \Vert g\Vert=\mu (E)$. This doesn't imply $f=0$ a.e.  Which functions $g$ should be chosen so that by (1) it is an upper bound of $\vert \int f\vert$ and tends $0$?
 A: Use $f$ to construct a densely defined functional $\delta:g\mapsto\int fg$ on $L^p$, $0<p<1$, that is bounded. Extend by continuity. 
Show that the space $L^p$, for $0<p<1$ doesn't have non-zero bounded linear functionals: A non-zero bounded linear functional $\delta$ produces non trivial open convex sets like $\{x:\delta(x)<1\}$. Show that the only non-empty convex open in $L^p$ is the whole space.
Therefore $\delta=0$. Hence $\int fg=0$ for all $g$ in $L^p$. What is left is the standard exercise in which you prove that $f=0$, a.e.

The standard proof of no trivial open sets:
It is enough to do the proof for $L^p[0,1]$.
Let $C\subset L^p[0,1]$ be a non-empty open convex set. 
By translation we can assume it contains the origin. Let $B_r\subset C$ be an open ball of radius $r$. Let $h\in L^p$ be any function. Let $N\in\mathbb{N}$ be large enough such that $\frac{d_p(h,0)}{N^{1-p}}<r$, where $d_p$ is the $L^p$ metric. 
Let $0=x_0<x_1<...<x_N=1$ such that $$\int_{x_i}^{x_{i+1}}|h|^p<\frac{d_p(h,0)}{N}$$
Define $g_i(t)=Nh(t)$ if $x_i\leq t<x_{i+1}$ and zero otherwise. Then $g_i\in B_r\subset C$ since $d_p(g_i,0)<\frac{d_p(h,0)}{N^{1-p}}<r$.
Since $C$ is convex $h=\frac{\sum_i g_i}{N}\in C$. Therefore $C=L^p[0,1]$.
