Suppose $f:\mathbb{R}\to\mathbb{R}$ is Borel measurable. I want to show that $\displaystyle \int_a^b f(x)\mathrm dx=0$ for all $-\infty
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4Consider the collection C of Borel sets B such that the integral of f on B is zero. You already know C contains every interval (a,b). What is the structure of C and what theorem from your course does this evoke? – 2011-08-24
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0Perhaps the Lebesgue Density Theorem? – 2011-08-24
1 Answers
If it's true of all rational $a
Later edit: Following a suggestion in the comments, I am adding this appendix to my answer. "Recall" (?) that:
The Lebesgue integral $\int f$ of any non-negative measurable function exists and is either $\infty$ or a finite number.
Any measurable function $f$ can be written as a sum $f^+ - f^-$ of positive and negative parts $f^+$ and $f^-$, where $f^+(x) = f(x)$ if $f(x) \ge 0$ and $f^+(x) = 0$ if $f(x) < 0$, and likewise $f^-(x)= -f(x) > 0$ if $f(x) < 0$ and $f^-(x) = 0$ if $f(x)>0$. The Lebesgue integral $\int f$ is then defined as $\int f^+ - \int f^-$ provided these are not both infinite. $f$ is said to be integrable precisely if both of these integrals are finite. That is the same as saying $\int|f| < \infty$.
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0Well, maybe it will be clearer if you add it in the answer. – 2011-08-24
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0And if it's true for reals $a\lt b$ the result can be derived like [this](http://math.stackexchange.com/a/158326/8271). – 2012-08-07