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Let $D(a,r)$ be an open ball in $\mathbb{R}^{k}$ ($ k\geq1 $), and $f$ locally integrable function in $\mathbb{R}^{k}$. Do we have: $$\lim_{r\to 0}\int_{D(a,r)}f(t)dt=0?$$

M. Rahmat
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    Yes, by dominated convergence. – Angina Seng Oct 26 '18 at 16:45
  • In physics there are cases where one uses point charges to describe fields. A point charge is represented by the Dirac delta function. If you integrate over a sphere around the point charge, and then shrink the radius to $+0$, what you get is a constant term, not necessarily zero. – M. Wind Oct 26 '18 at 18:02
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    The question says explicitly "$f$ is a locally integrable function" ruling out the possibility that $f$ is a Dirac's delta. – Julián Aguirre Oct 26 '18 at 18:21
  • Lord Shark the Unknown: How would you deduce this from dominated convergence? I don't see it. – M. Rahmat Oct 27 '18 at 04:54
  • Why would it not be 0? Out of all the numbers out there, such as: 1, 15, 0, 92, 5673, -14i, why would it be anything other than 0 ? – Nike Dattani Oct 28 '18 at 22:08

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The answer is YES. This is a classic result called "absolute continuity of The Lebesgue integral", for example see rmh52 (https://math.stackexchange.com/users/35961/rmh52), Absolute continuity of the Lebesgue integral, URL (version: 2017-02-03): Absolute continuity of the Lebesgue integral

M. Rahmat
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