Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f \in L^1([0,1])$ be such that for all $t \geq s$, $\displaystyle \int_s^t f(u)du \leq 0$. Is it true that $f\leq 0$ almost everywhere?

share|cite|improve this question
up vote 1 down vote accepted

We will show that if $f$ is integrable and $\int_{(s,t)}f(u)du\geq 0$ for all $s<t$ then $f\geq 0$ almost everywhere. Let $A\subset [0,1]$ Borel measurable and $\varepsilon>0$. As $f$ is integrable, we can find $\delta>0$ such that is $B\subset [0,1]$ is measurable and $\lambda(B)\leq \delta$ then $\left|\int_Bf(u)du\right|\leq \varepsilon$. By regularity of $\lambda$, let $O$ open such that $A\subset O$ and $\lambda(O\setminus A)\leq\delta$. We can write $O=\bigsqcup_{j\geq 1}I_{j}$, where $I_j$ are pairwise disjoint intervals. This gives $$\int_A fd\lambda=\int_Ofd\lambda+\int_{O\setminus A}fd\lambda\geq \sum_{j\geq 1} \int_{I_j}fd\lambda-\varepsilon\geq -\varepsilon.$$ As $\varepsilon$ is arbitrary, we have for all $A$ Borel measurable that $\int_A fd\lambda\geq 0$. Now we apply this to $A=\{x, g(x)<-1/n\}$ for $g$ representing $f$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.