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be $G\subseteq \mathbb{R}$ a open set. Be $f:G\rightarrow \mathbb{R}$* measured and for all interval $[a,b] \subseteq G$ to have that $f$ is lebesgue integrable function in $[a,b]$ and $\int_{a}^b f dm=0$. Show that $f=0$ almost everywhere.

I know that if for all set measured $A\subseteq E$, if $\int_{A}f=0$ so $f=0$ almost everywhere, I try to use this for the problem but dont work.

thanks

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Hint: Which structure has the collection of the Borel subsets $A$ of $\mathbb R$ such that the integral of $f$ on $A$ is zero? –  Did Jun 13 '12 at 22:54
    
I think the borel set $F_{\delta}$ –  kEoz Jun 14 '12 at 0:14
    
is $\mathrm{d}m$ Lebesgue measure? –  robjohn Jul 8 '12 at 0:20

3 Answers 3

$\newcommand{\d}{\ \mathrm d}$ Another way to approach this is by using the Lebesgue differentiation theorem (LDT).

Since $G$ is open, given $x\in G$, there is a $\delta\gt 0$ such that $(x-\delta, x+\delta)\subseteq G$. This says that for any $\delta\gt 0$, small enough, we can consider $\int_{x-\delta}^{x+\delta} f(t)\d m(t)$. Therefore $$\begin{align*} f(x) &= \lim_{\delta\to 0} \frac1{2\delta}\int_{x-\delta}^{x+\delta} f(t)\d m(t)\quad\text{a.e.} &&\text{by LDT}\\ &= \lim_{\delta\to 0} \frac1{2\delta}\cdot 0\quad\text{a.e.} &&\text{by your hypothesis}\\ &= 0\quad\text{a.e.} \end{align*}$$ as we wanted.

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The Lebesgue measure is based on the borelian $\sigma$-algebra on $\mathbb{R}$. That means that measurable sets come as a limit of open and closed sets in terms of unions and intersections. That implies that for every measurable set $A$, there is a sequence of unions of intervals $I_n$ such that $\chi_{I_n}\to \chi_A$ pointwise. Because: $$ \int_{I_n}fd\mu=\int \chi_{I_n}fd\mu $$ then in each $A$ where the integral makes sense, there will be a 0 limit. And then you can use your argument. That works also for functions that are not necessarily positive.

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Suppose $f(x) > \epsilon >0$ on some $A \subseteq G$ such that $\mu (A) >0.$ Since $G$ is open, there exists open $U$ with $A \subseteq U \subseteq G.$ Note that $U$ can be written as the countable union of disjoint intervals $\displaystyle\bigcup_{n\in \mathbb{N}} I_n$ with each $I_n \subseteq G.$ Hence $\displaystyle\sum_{n=1}^{\infty} \displaystyle\int_{I_n} f d\mu = \displaystyle\int_U f d\mu \ge \displaystyle\int_A f d\mu > \epsilon \cdot \mu (A) >0,$ a contradiction.

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you sure the the existence of that interval? –  kEoz Jun 13 '12 at 22:53
    
No, in fact I was wrong. Check the edit. –  user17794 Jun 13 '12 at 23:09
    
with your argument I just write G like union disjoint intervals,and what type of intervals you take? –  kEoz Jun 14 '12 at 0:15
    
Well, my argument writes $U$ as countable union of disjoint intervals. The sum on the left has to be zero by your assumption. –  user17794 Jun 14 '12 at 4:31
    
I noticed you haven't accepted any answers to the questions you've posted. Without intimating that you ought to accept mine (it is less elegant than the others), it is MSE custom to accept an answer that you feel has sufficiently answered your question. –  user17794 Jun 15 '12 at 19:01

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