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 $\mu(\{f > 0\}) > 1 - \varepsilon$?



Is this function positive on a set of large measure?
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.


