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.
If there is no $\delta$ corresponding to some $\epsilon$ then $\mu (|f_n| >\epsilon)\to 0$ for some $(f_n) \subseteq E_p$ But then $1=\int fd\mu \leq \epsilon +\int_{|f_n| >\epsilon}|f_n|d\mu$. We can reach a contradiction using uniform integrability: the fact that $\int |f_n|^{p} d\mu$ is bounded implies that $(f_n)$ is uniformly integrable and hence $\sup_n \int_A |f_n| d\mu \to 0$ as $\mu (A) \to 0$.
Reference for uniform integrability: https://www.randomservices.org/random/expect/Uniform.html