Throughout, we denote by $\mu$ the Lebesgue measure on $[0, 1]$.
Let $f \in L^1([0, 1])$ be a nowhere zero function, that is, the set of all $x \in [0, 1]$ such that $f(x) = 0$ is empty.
Suppose there exists some $\varepsilon$ with $0 < \varepsilon < 1$ such that for every open subset $E$ of $[0, 1]$ with Lebesgue measure $\varepsilon$, we have
$$\int_E f \, d\mu > 0.$$
Question: Is it true that $\mu(\{f > 0\}) > 1 - \varepsilon$?
I'm guessing your question means what it says literally, that $\int_U f>0$ for every open set $U$ of measure exactly $\epsilon$. Is this right? (The answer even with this restriction is still positive).
Let $N=\{x\colon f(x)<0\}$ and suppose for a contradiction that $\mu(N)\ge\epsilon$. Let $N_1$ be a subset of $N$ of measure exactly $\epsilon$ and let $\delta<\frac 12\int_{N_1}|f|\,d\mu$. Let $0<\eta<\epsilon$ be such that $\int_A |f|<\delta$ whenever $\mu(A)\le\eta$. Now let $N_2$ be a measurable subset of $N_1$ of measure $\epsilon-\eta$. By regularity of Lebesgue measure, there exists an open set $U$ containing $N_2$ of measure at most $\epsilon$. By the choice of $\eta$, we see $\int_{U}f<-\delta$.
Now let $s$ be chosen so that $\mu([0,s)\setminus U)=\epsilon-\mu(U)$ (so that in particular, $\mu([0,s)\setminus U)<\eta$. Now we see $\mu([0,s)\cup U)=\epsilon$ and $\int_{[0,s)\cup U}f<0$, giving the required contradiction.
