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