# The Unapologetic Mathematician

## 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.

December 15, 2009 - Posted by | Analysis, Calculus

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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 […]

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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 […]

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