The Unapologetic Mathematician

Mathematics for the interested outsider

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 f:[a,b]\rightarrow\mathbb{R} defined on an n-dimensional interval [a,b] is Riemann integrable on that interval if and only if the set D of discontinuities of f 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

\displaystyle J_\epsilon=\{x\in[a,b]\vert\omega_f(x)\geq\epsilon\}

and by our earlier condition we know that \overline{c}(J_\epsilon)=0 for all \epsilon>0. In particular, it holds for \epsilon=\frac{1}{n} for all natural numbers n.

If x\in D is a point where f is discontinuous, then the oscillation \omega_f(x) must be nonzero, and so \omega_f(x)>\frac{1}{n} for some n. That is

\displaystyle D\subseteq\bigcup\limits_{n=1}^\infty J_{\frac{1}{n}}

Since \overline{c}(J_{\frac{1}{n}})=0, we also have \overline{m}(J_{\frac{1}{n}})=0, and therefore have \overline{m}(D)=0 as well.

Conversely, let’s assume that \overline{m}(D)=0. Given an \epsilon>0, we know that J_\epsilon is a closed set contained in D. From this, we conclude that \overline{c}(J_\epsilon)\leq\overline{m}(D)=0. Since this is true for all \epsilon, the Jordan content condition holds, and f is Riemann integrable.

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December 15, 2009 - Posted by | Analysis, Calculus

4 Comments »

  1. [...] 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 | Reply

  2. [...] 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 [...]

    Pingback by Integrals Over More General Sets « The Unapologetic Mathematician | December 22, 2009 | Reply

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

    Pingback by Integrals are Additive Over Regions « The Unapologetic Mathematician | December 30, 2009 | Reply

  4. [...] 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 [...]

    Pingback by Almost Everywhere « The Unapologetic Mathematician | May 13, 2010 | Reply


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