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Exercise 1.6.14 (Locally integrable functions)
Show that is locally integrable if and only if
Show that Theorem 1.6.19 implies a generalisation of itself in which the condition of absolute integrability of is weakened to local integrability.
Answers
We demonstrate that both definitions of the local absolute integrability are indeed equivalent.
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Suppose that for any we can find an open neighbourhood of on which the integral of is finite. The trick is to notice how this condition implies that is absolutely integrable on every compact set in . To see why, let be an arbitrary compact set. For each we can find an open set on which is absolutely integrable. But obviously, is an open cover for , and so we can find a finite subcover from such than covers . By additivity of the Lebesgue integral, we then obtain
Now consider the ball for arbitrary radius . The boundary of this ball is a null set, and so
since is compact by Heine-Borel theorem.
- Suppose that for any . Pick an arbitrary . We cannot use since it is not centred at . The trick is to look at on instead. It is easy to verify using the triangle inequality that , and so we have found a neighbourhood of at which it is guaranteed that . Since our choice of was arbitrary, the claim follows.
This result can be used to generalise Theorem 1.6.19 (Lebesgue differentiation theorem in general dimension) to any locally integrable function . This is done by considering a point , finding a open neighbourhood of on which is absolutely integrable. We then apply Theorem 1.6.19 to the restriction (one would of course choose small enough so that the balls in the limit pass inside the ).