The Lebesgue Integral of Bounded Riemann Integrable Functions

# The Lebesgue Integral of Bounded Riemann Integrable Functions

Recall from The Lebesgue Integral of Bounded Functions page that if $f$ is a bounded function defined on a Lebesgue measurable set $E$ with $m(E)$ then $f$ is said to be Lebesgue measurable if its upper and lower Lebesgue integrals are equal, that is:

(1)
\begin{align} \quad (L) \overline{\int_E} f = \inf \left \{ \int_E \psi : \psi \: \mathrm{is \: a \: simple \: function}, \: \psi(x) \geq f(x) \: \mathrm{on \:} E \right \} = \sup \left \{ \int_E \varphi : \varphi \: \mathrm{is \: a \: simple \: function}, \varphi(x) \leq f(x) \: \mathrm{on \:} E \right \} = (L) \underline{\int_E} f \end{align}

We now show that if $f$ is a bounded, Riemann integrable function on the closed and bounded interval $[a, b]$ then $f$ is also Lebesgue integrable on $[a, b]$ and the Lebesgue integral of $f$ on $[a, b]$ equals the Riemann integral of $f$ on $[a, b]$.

 Theorem 1: Let $f$ be a bounded, Riemann integrable function on $[a, b]$. Then $f$ is Lebesgue integrable on $[a, b]$ and $\displaystyle{(L) \int_{[a, b]} f = (R) \int_a^b f}$.
• Proof: We use the fact that every step function is a simple function. So:
(2)
\begin{align} \quad (R) \overline{\int_a^b} f &= \inf \left \{ (R) \int_a^b \psi : \psi \: \mathrm{is \: a \: step \: function}, \: \psi(x) \geq f(x) \: \mathrm{on \:} [a, b] \right \} \\ & \geq \inf \left \{ (L) \int_{[a, b]} \psi : \psi \: \mathrm{is \: a \: simple \: function}, \: \psi(x) \geq f(x) \: \mathrm{on \:} [a, b] \right \} \\ & \geq (L) \overline{\int_{[a, b]}} f \\ & \geq (L) \underline{\int_{[a, b]}} f \\ & \geq \sup \left \{ (L) \int_{[a, b]} \varphi : \varphi \: \mathrm{is \: a \: simple \: function}, \: \varphi(x) \leq f(x) \: \mathrm{on \:} [a, b] \right \} \\ & \geq \sup \left \{ (R) \int_{[a, b]} \varphi : \varphi \: \mathrm{is \: a \: step \: function}, \: \varphi(x) \leq f(x) \: \mathrm{on \:} [a, b] \right \} = (R) \underline{\int_a^b} f \end{align}
• Since $f$ is Riemann integrable on $[a, b]$ we have that $\displaystyle{(R) \overline{\int_a^b} f = (R) \underline{\int_a^b} f}$. So from the inequality above we have that:
(3)
\begin{align} \quad (L) \overline{\int_{[a, b]}} f = (L) \underline{\int_{[a, b]}} f \end{align}
• So $f$ is Lebesgue integrable on $E$. $\blacksquare$