# The Lebesgue Criterion for Riemann Integrability — a proof without using the concept of oscillation.

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: math.stackexchange.com/a/163409/442 –  GEdgar Nov 25 '12 at 22:46