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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?

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  • $\begingroup$ You need to follow your intuition and make it concrete. $\endgroup$ – copper.hat Aug 16 '18 at 17:45
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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 $\int_{A_2} f \le \int _{A_3} f$.

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  • $\begingroup$ I don't see why $\mu(A_2)=\mu(A_3).$ Can you give me a hint? $\endgroup$ – mfl Aug 16 '18 at 17:57
  • $\begingroup$ I see that: $\mu(A_2)+\mu(E)=\mu(A_2\cup E)=\mu([0,t]\cup E)=\mu(A_3\cup E)=\mu(A_3)+\mu(E)$. $\endgroup$ – mathstackuser Aug 16 '18 at 17:59
  • $\begingroup$ 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? $\endgroup$ – mathstackuser Aug 16 '18 at 18:01
  • $\begingroup$ $\mu A_1+ \mu A_2 = \mu [0,t] = t = \mu E = \mu A_1 + \mu A_3$. $\endgroup$ – copper.hat Aug 16 '18 at 18:01
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    $\begingroup$ @copper.hat Thank you for the clarification. $\endgroup$ – mfl Aug 16 '18 at 18:03
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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 $f\not \in L^1$.

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Hint

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

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  • $\begingroup$ I tried, but I have difficulties to deal with that. Could you give some more advise please? $\endgroup$ – mathstackuser Aug 16 '18 at 17:39
  • $\begingroup$ 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 $f=\chi_I$. $\endgroup$ – David C. Ullrich Aug 16 '18 at 18:46

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