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

# Corollaries to Lebesgue's Criterion for the Riemann Integrability of Bounded Functions

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 $g$ on $[a, b]$ 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 $[a, b]$. $\blacksquare$