## Lebesgue’s Condition

At last we come to Lebesgue’s condition for Riemann-integrability in terms of Lebesgue measure. It asserts, simply enough, that a bounded function defined on an -dimensional interval is Riemann integrable on that interval if and only if the set of discontinuities of has measure zero. Our proof will go proceed by way of our condition in terms of Jordan content.

As in our proof of this latter condition, we define

and by our earlier condition we know that for all . In particular, it holds for for all natural numbers .

If is a point where is discontinuous, then the oscillation must be nonzero, and so for some . That is

Since , we also have , and therefore have as well.

Conversely, let’s assume that . Given an , we know that is a closed set contained in . From this, we conclude that . Since this is true for all , the Jordan content condition holds, and is Riemann integrable.

[...] it has outer Lebesgue measure zero in the rectangle . If these are the only discontinuities, then is integrable on , but we can’t follow the above prescription anymore, even if it were actually rigorous. [...]

Pingback by Iterated Integrals I « The Unapologetic Mathematician | December 16, 2009 |

[...] be integrable over if and only if the discontinuities of in form a set of measure zero. Indeed, Lebesgue’s condition tells us that the discontinuities of must have measure zero. These discontinuities either come [...]

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[...] can use Lebesgue’s condition to establish our assertion about integrability. On the one hand, the discontinuities of in must [...]

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[...] actually seen this sort of thing in the wild before; Lebesgue’s condition can be reformulated to say that a bounded function defined on an -dimensional interval is Riemann [...]

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