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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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    is $\mathrm{d}m$ Lebesgue measure?2012-07-08

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$\newcommand{\d}{\ \mathrm d}$ 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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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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    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.2012-06-15
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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.