Corollaries to Lebesgue's Crit. - Riemann Integrability of Bnd. Funct.

Table of Contents

Recall from the Lebesgue's Criterion Part 1 - Riemann Integrability of a Bounded Function and Lebesgue's Criterion Part 2 - Riemann Integrability of a Bounded Function pages that if $$f$$ is a bounded function on $$[a, b]$$ and if $$D$$ is the set of discontinuities of $$f$$ on $$D$$ then $$f$$ is Riemann integrable on $$[a, b]$$ if and only if the set of discontinuities has measure $$0$$ , i.e., $$m(D) = 0$$ .

We will now state some very nice corollaries that come as a consequence of Lebesgue's criterion.

Corollary 1: Let

$f$

be a function of bounded variation on

$[a, b]$

. Then

$f$

is Riemann integrable on

$[a, b]$

Proof: If $$f$$ is a function of bounded variation on $$[a, b]$$ then we have already proven that $$f$$ is bounded on $$[a, b]$$ on the Functions of Bounded Variation page.

Furthermore, from the Decomposition of Functions of Bounded Variation as the Difference of Two Increasing Functions page we have that $$f$$ can be written as the difference of two increasing functions, say $f = \alpha_1 - \alpha_2$ where $\alpha_1, \alpha_2$ are both increasing on $$[a, b]$$ .

From the Countable Discontinuities of Monotonic Functions page we know that $\alpha_1$ and $\alpha_2$ both have countably many discontinuities, and so the set $$D$$ of all discontinuties of $$f$$ is countable.

From The Measure of Countable Subsets of Real Numbers page we see that since $$D$$ is countable that then $$m(D) = 0$$ . So, by Lebesgue's Criterion we see that $$f$$ is Riemann integrable on $$[a, b]$$ . $\blacksquare$

Corollary 2: Let

$f$

and

$g$

be bounded functions on

$[a, b]$

. Let

$D_f$

and

$D_g$

denote the set of discontinuities of

$f$

and

$g$

$[a, b]$

respectively. Suppose that

$D_f = D_g$

. Then

$f$

is Riemann integrable on

$[a, b]$

if and only if

$g$

is Riemann integrable on

$[a, b]$

Proof: $\Rightarrow$ Suppose that $$f$$ is Riemann integrable on $$[a, b]$$ . Then $$m(D_f) = 0$$ . But since $$D_f = D_g$$ we have that $$m(D_g) = 0$$ . So $$g$$ is Riemann integrable on $$[a, b]$$ .

$\Leftarrow$ Similarly, suppose that $$g$$ is Riemann integrable on $$[a, b]$$ . Then $$m(D_g) = 0$$ . But since $$D_f = D_g$$ we have that $$m(D_f) = 0$$ . So $$f$$ is Riemann integrable on $$[a, b]$$ . $\blacksquare$

Mathonline