Let $E$ be a subset of $\Bbb R$ with positive Lebesgue measure, $\lambda(E)>0$. Let $f$ be a function from $\Bbb R$ to $\Bbb R$ which is positive on $E$, that is $f(x)>0$ for all $x\in E$.

Is it possible that $$\int_E f\,d\lambda=0?$$ In other words, must $$\int_E f\,d\lambda$$ be strictly positive?

  • 4
    $\begingroup$ $\int_E f\,d\lambda=0 \iff f=0$ almost everywhere $\endgroup$ – iwriteonbananas Jan 28 '15 at 19:53
  • 1
    $\begingroup$ you can also approximate $f$ by simple functions. On a subset of $F\subset E$ you will find that $f_k\leq f$ and $f_k\geq \varepsilon >0$ $\endgroup$ – Quickbeam2k1 Jan 28 '15 at 20:05
  • $\begingroup$ Does anyone know of a name for this result? I would like to be able to write something like "... so f is zero a.e. by the such-and-such lemma..." within the context of a larger proof. $\endgroup$ – Nick Alger May 11 '20 at 7:52

Since $f$ is strictly positive on $E$, we have $$ E = \bigcup_{n \geq 1} E_n, \quad \mbox{ where } E_n = \left\{x \in E: f(x) > \frac{1}{n}\right\}. $$ Since $\lambda(E) > 0$ there is some $n$ for which $\lambda(E_n)$ is positive (otherwise $E$ would be the countable union of measure $0$ sets, implying $\lambda(E)=0$). We then have $$ \int_E f \, d\lambda \geq \int_{E_n} f \, d\lambda > \int_{E_n} \frac1n \, d\lambda = \frac{\lambda(E_n)}{n} > 0, $$ as desired.


Could I create a simple function $K(x)$, where $$K(x)=\begin{cases}1,\text{ if } x\notin E\\0,\text{ if } x\in E\end{cases}\quad ?$$

If $f(x)>0$, this would imply that on the set $E$, $f(x)>K(x),\space\forall x\in E$.

Further by the monotonicity nature of the Lebesgue integral, the integral of $\int f(x)dx>\int K(x)dy=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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