Exercise 1.6.14 (Locally integrable functions)

Show that f : d is locally integrable if and only if

r > 0 : B(0,r)|f|dm < .

Show that Theorem 1.6.19 implies a generalisation of itself in which the condition of absolute integrability of f is weakened to local integrability.

Answers

We demonstrate that both definitions of the local absolute integrability are indeed equivalent.

  • Suppose that for any x d we can find an open neighbourhood of x on which the integral of f is finite. The trick is to notice how this condition implies that f is absolutely integrable on every compact set in Rn. To see why, let K d be an arbitrary compact set. For each x K we can find an open set Ox on which f is absolutely integrable. But obviously, xKOk is an open cover for K, and so we can find a finite subcover O1,,On from (Ox)xK such than i=1nOi covers K. By additivity of the Lebesgue integral, we then obtain

    K|f|dm i=1nOi|f|dm < .

    Now consider the ball B(0,r) for arbitrary radius r > 0. The boundary of this ball ∂B(0,r) is a null set, and so

    B(0,r)|f|dm =B(0,r)¯|f|dm 0 <

    since B(0,r)¯ is compact by Heine-Borel theorem.

  • Suppose that B(0,r)|f| < for any r > 0. Pick an arbitrary x d. We cannot use B(0,r) since it is not centred at x. The trick is to look at B(0,2x)|f| < on B(0,2x) instead. It is easy to verify using the triangle inequality that B(x,x) B(0,2x), and so we have found a neighbourhood B(x,x) of x at which it is guaranteed that B(x,x)|f| < . Since our choice of x d 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 f. This is done by considering a point x d, finding a open neighbourhood U of d on which f is absolutely integrable. We then apply Theorem 1.6.19 to the restriction f U (one would of course choose r > 0 small enough so that the balls B(0,r) in the limit pass inside the U).

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2021-03-28 00:00
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