Let $E\subset [0,1]$ with $\mu(E)=t$ and $f$ an increasing function. Prove that $\int_0^t f\leq \int_E f$.

The claim seems to be pretty clear to me and I can imagine why this has to be true but I have a hard time to prove it. Any hints on that?

  • You need to follow your intuition and make it concrete. – copper.hat Aug 16 at 17:45

Hints:

Let $A_1=[0,t] \cap E$, $A_2 = [0,t] \setminus E$ and $A_3 = E \setminus [0,t]$.

Note that $\mu A_2 = \mu A_3$ and $f(x_2) \le f(t)$ for all $x_2 \in A_2$ and $f(t) \le f(x_3)$ for all $x_3 \in A_3$.

You need to show that $\int_{A_2} f \le \int _{A_3} f$.

  • I don't see why $\mu(A_2)=\mu(A_3).$ Can you give me a hint? – mfl Aug 16 at 17:57
  • I see that: $\mu(A_2)+\mu(E)=\mu(A_2\cup E)=\mu([0,t]\cup E)=\mu(A_3\cup E)=\mu(A_3)+\mu(E)$. – mathstackuser Aug 16 at 17:59
  • But how can I show that $\int_{A_2}f\leq\int_{A_3} f$? I see how to use it after I showed that. Where do you need $A_1$ for? – mathstackuser Aug 16 at 18:01
  • $\mu A_1+ \mu A_2 = \mu [0,t] = t = \mu E = \mu A_1 + \mu A_3$. – copper.hat Aug 16 at 18:01
  • 1
    @copper.hat Thank you for the clarification. – mfl Aug 16 at 18:03

This result would be obvious if $E$ was an interval, or a finite union of intervals. To extend this to general $E$, use the fact that any measurable set can be approximated by a finite union of intervals. Namely, for any $\delta>0$, there exists a set $F$ so that $F$ is a finite union of intervals and $\mu(E\oplus F)<\delta$, where $E\oplus F=(E\setminus F) \cup (F\setminus E)$ is the symmetric difference of $E$ and $F$.

You then have $$ \Big|\int_E f\,d\mu-\int_F fd\mu\Big|\le \int_{E\oplus F}|f| $$ Now, if $f\in L_1$, then the above integral would become arbitrarily small as $\mu(E\oplus F)\to 0$. (The result I am using is $f\in L^1$ implies for all $\epsilon>0$ that there exists a $\delta$ so $\mu(A)<\delta$ implies $\int_A |f|\,d\mu<\epsilon$, which you should try to prove.) Since $|\int_E f\,\mu -\int_F f\,d\mu|$ is arbitrarily small, and $\int_F f\,d\mu \ge \int_0^t f\,d\mu,$ you can show that $\int_E f\,d\mu \ge \int_0^t f\,d\mu$.

I leave it to you to figure out what to do $f\not \in L^1$.

Hint

Use simple functions that approximate $f$ on $[0,1]$.

  • I tried, but I have difficulties to deal with that. Could you give some more advise please? – mathstackuser Aug 16 at 17:39
  • If $f$ is non-decreasing then it can be uniformly approximated by $\sum c_j\chi_{I_j}$, where each $I_j$ is of the form $(a,1]$ or $[a,1]$ and $c_j>0$. So you can assume that $f=\chi_I$. – David C. Ullrich Aug 16 at 18:46

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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