Any help with this problem is appreciated.

Given the $f$ is measurable and finite a.e. on $[0,1]$. Then prove the following statements

$$ \int_E f = 0 \text{ for all measurable $E \subset [0,1]$ with $\mu(E) = 1/2$ }\Rightarrow f = 0 \text{ a.e. on } [0,1]$$ $$ f > 0 \text{ a.e. } \Rightarrow \inf ~ \left\{\int_E f : \mu(E) \geq 1/2\right\} > 0 $$

  • $\begingroup$ I am confused as to how does the measure of set $E$ come into the picture. $\endgroup$ – user62089 Feb 24 '13 at 19:22
  • $\begingroup$ What did you try? $\endgroup$ – Did Feb 24 '13 at 19:33
  • $\begingroup$ I have no idea where to start. I was just trying to see when is $\mu(E)$ important in the question $\endgroup$ – user62089 Feb 24 '13 at 19:34
  • $\begingroup$ Can you show that the integral of $f$ on $(1/4,1/2)$ is zero? $\endgroup$ – Did Feb 24 '13 at 19:41
  • 2
    $\begingroup$ @pondy You are not following my suggestions. Let me try once again: let $A_i$ denote the integral of $f$ on $((i-1)/4,i/4)$, then $A_1+A_2=0$ (why?), $A_2+A_3=0$ (why?) and $A_1+A_3=0$ (why?) hence $A_2=0$ (why?) hence $A_i=0$ for every $i$ (why?). Then generalize. $\endgroup$ – Did Feb 25 '13 at 7:20

For $(1)$, we can define the sets $ P:= \{x:f(x)\ge 0\}$ and $ N:=\{x:f(x)\le 0\}$. Then either $\mu(P)\ge \frac{1}{2}$ or $\mu(N)\ge \frac{1}{2}$. Suppose $\mu(P)\ge \frac{1}{2}$, define $SP:= \{x:f(x)> 0\},$ then $\ SP\subset P$. If $ \mu(SP)<\frac{1}{2}$, we can choose a set $E$ such that $$SP \subset E,\ f(x)\ge 0\ on \ E,\ and\ \mu(E)=\frac{1}{2}$$ According to the hypothesis, $\int_Ef=0$ which implies $\mu(SP)=0$($f$ is non-negative on $E\ \Rightarrow \ f=0\ a.e.\ on\ E$). I think you are able to show the case when $\mu(SP)>\frac{1}{2}$. Similarly, if we define $ SN:=\{x: f(x)<0\} $, we can show $\mu(SN)=0$.

For $(2)$, forst define $A_n:=\{x:f(x)>\frac{1}{n}\} $, then we know $A_n$ is incresing and $\lim_{n\to \infty}\mu(A_n)=1$ since $f>0\ a.e.$. Fix sufficiently large $n_0$ so that $\mu(A_{n_0})>1-\epsilon_0$, then for any $E$ with $\mu(E)\ge \frac{1}{2}$, we have $$ \int_Ef\,\mathrm{d}\mu=\int_{E\cap A_{n_0}^c} f\,\mathrm{d}\mu+\int_{E\cap A_{n_0}} f\,\mathrm{d}\mu\ge \int_{E\cap A_{n_0}} f\,\mathrm{d}\mu\ge \frac{1}{n_o}\cdot \mu(E\cap A_{n_0})$$ Note that $\mu(E\cap A_{n_0})\ge \mu (E)+\mu (A_{n_0})-1> \frac{1}{2}+(1-\epsilon_0)-1=\frac{1}{2}-\epsilon_0$, hence $\int_E f\,\mathrm{d}\mu>\frac{1}{n_0}\cdot (\frac{1}{2}-\epsilon_0)>0$.


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.