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I am trying to prove the Lebesgue Criterion for Riemann Integrability without using the concept of oscillation.

The Lebesgue Criterion for Riemann Integrability states that if $ f: [a,b] \to \mathbb{R} $ is bounded, then $ f $ is Riemann integrable iff the set of discontinuities of $ f $ has measure $ 0 $.

This is what I have so far:

Firstly, define $$ D \stackrel{\text{df}}{=} \{ x \in [a,b] \mid \text{$ f $ is not continuous at $ x $} \}. $$ Next, for each $ \epsilon > 0 $ and each $ n \in \mathbb{N} $, define $ A(\epsilon,n) $ to be the union of those intervals $ I $ that satisfy the following conditions:

  • $ I $ is a member of the regular $ 2^{n} $-partition of $ [a,b] $.
  • $ \displaystyle \sup_{x \in I} f(x) - \inf_{x \in I} f(x) > \epsilon $.

Now, note that $$ x \in D \quad \iff \quad (\forall \epsilon > 0)(\exists N \in \mathbb{N})(\forall n \in \mathbb{N}_{\geq N}) (x \in A(\epsilon,n)). $$ Therefore, we can say that $$ D = \bigcup_{\epsilon > 0} \left[ \bigcap_{n \in \mathbb{N}} A(\epsilon,n) \right] = \bigcup_{k \in \mathbb{N}} \left[ \bigcap_{n \in \mathbb{N}} A \! \left( \frac{1}{k},n \right) \right]. $$ From here, I am lost.

Any help would be greatly appreciated. Thank you.

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I wrote an answer for the case with values in Banach space here: – GEdgar Nov 25 '12 at 22:46

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