Let $f$ be a Lebesgue measurable function on $\Bbb{R}$ satisfying some properties, prove $f\equiv 0$ a.e. 
Let $f$ be a Lebesgue measurable function on $\Bbb{R}$ satisfying:
i) there is $p\in (1,\infty)$ such that $f\in L^p(I)$ for any bounded interval $I$.
ii) there is some $\theta \in (0,1)$ such that: $$\left|\int_I f\ dx\right|^p\leq \theta (\mu(I))^{p-1}\int_I |f|^p\ dx$$
Prove that $f\equiv 0 $ a.e.

This is a previous qual question (3, beware: PDF).
My thoughts.
We assume that there is some $E$ where without loss of generality $f>0$ on $E$ and $\mu(E)>0$. Using regularity of Lebesgue measure, for all $\epsilon>0$ there is some open $G_{\epsilon}$ with $\mu(G_{\epsilon}\setminus E)<\epsilon$ and $E \subseteq G_{\epsilon}$. We may decompose $G_{\epsilon}$ as a countable disjoint union $I_k$. So we must only show that there is a contradiction if $f>0$ on some interval.
I tried arguing like this:
$$\begin{align*}\left|\int_I f\ dx\right|^p&\leq \theta (\mu(I))^{p-1}\int_I |f|^p\ dx\\ &\leq\theta (\mu(I))^{p-1} \mu(I) \operatorname{essup}(|f|^p)\\&=\theta(\mu(I))^p\operatorname{essup}_I(|f|^p)\end{align*}$$
I feel like there should be a general contradiction here (for example if $I=(0,1)$ and $f=1$ this inequality doesn't hold).
Can someone help me? I would like hints only please.

Hint: Suppose $\theta = 1/2$ just to scratch around a bit. For $h>0$ we have
$$|\int_a^{a+h} f\,\,|^p \le \frac{1}{2}\cdot h^{p-1}\int_a^{a+h} |f|^p \implies  |\frac{1}{h} \int_a^{a+h} f|^p \le \frac{1}{2}\cdot \frac{1}{h}\int_a^{a+h} |f|^p.$$
In the last inequality, let $h\to 0^+$ and apply the Lebesgue differentiation theorem.
