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This question was asked in my exam yesterday and this is what I did.Will I get marks here?

Let $ f : \mathbb R \rightarrow \mathbb R$ be a bounded Lebesgue measurable function such that $\int_a^b f =0$ for all real $a,b.$

Show that $\int _E f=0$ for each subset $E $ of $\mathbb R $ of finite Lebesgue measure.

Since $m(E)<\infty$ so given $\epsilon>0$ there exists an open set $O\supseteq E$ such that $m(E)-m(O)<\epsilon.$ Now $E$ is measurable $\implies m(E\setminus O)<\epsilon$.

Also $O$ can be expressed as disjoint union of open intervals $O=\cup_{n=1}^\infty (a_n,b_n)$.

By Dominated Convergence Theorem, $\int _O f=\sum _{n=1}^\infty \int_{(a_n,b_n)} f$ and hence $\int _O f=0$

Hence $\int _O f=\int _E f+\int _{E\setminus O} f\implies \int _E f=0$ since $m(E\setminus O)=0$.

Please check my proof.

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    Why do you say $m(E\setminus O)=0$? – 2017-01-26
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    Because $m(E\setminus O)<\epsilon $ for all $\epsilon>0$ – 2017-01-26
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    But $O$ depends on $\epsilon$ so we can't take $\epsilon$ arbitrarily. – 2017-01-26
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    Will you please do the correct one? – 2017-01-26
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    Another issue is that you (correctly) write that $O\supset E$ and then suddenly you consider $E\setminus O$ (which is empty) and $m(E)-m(O)$ which is negative. – 2017-01-26

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