Question: Let $(X,A,m)$ be a measure space such that $m(X)=1$. For each $1<p<\infty$ define the set $E_p=\{f\in L^1(m):\int |f|dm=1 \text{ and} \int |f|^pdm=2\}$. Show that for each $0<\epsilon<1$ there exists some $\delta_{p,\epsilon}>0$ such that $m(\{x\in X:|f(x)|>\epsilon\})\geq\delta_{p,\epsilon}$ for each $f\in E_p$.
My Thoughts: So, I am reading what we are trying to prove as "Show that $f$ does not converge to $0$ in measure for each $f\in L^1$. But, since in our definition of $E_p$, can't we say that instead of $f$ just being in $L^1$, it is also in $L^p$ for $1\leq p<\infty$? So, I'd like to say $$m(|f|>\epsilon)= m(|f|>\epsilon, |f|>\epsilon^p)+m(|f|>\epsilon, |f|\leq\epsilon^p)$$ and then somehow use the integral values we have in the definition of $E_p$ to show that for $0<\epsilon<1$ we get the result. I'm just not sure if I have the right idea here, or if there is a different way I should be looking at it. Any insight, ideas, etc. are always greatly appreciated! Thank you.