The Lebesgue Integral of Bounded Lebesgue Measurable Functions

# The Lebesgue Integral of Bounded Lebesgue Measurable 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 function that is also a Lebesgue measurable function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$ then $f$ is Lebesgue integrable on $E$.

 Theorem 1: Let $f$ be a bounded, Lebesgue measurable function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. Then $f$ is Lebesgue integrable on $E$.
• Proof: Let $f$ be a bounded, Lebesgue measurable function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$.
• Let $\varphi$ and $\psi$ be step functions such that $\varphi(x) \leq f(x) \leq \psi(x)$ for all $x \in E$. Then clearly:
(2)
\begin{align} \quad (L) \underline{\int_E} f = \sup \left \{ \int_E \varphi : \varphi \: \mathrm{is \: a \: simple \: function}, \: \varphi(x) \leq f(x) \: \mathrm{on} \: E \right \} \leq \inf \left \{ \int_E \psi : \psi \: \mathrm{is \: a \: simple \: function}, \: \psi(x) \geq f(x) \: \mathrm{on} \: E \right \} = (L) \overline{\int_E} f \quad (*) \end{align}
• Now, let $\epsilon > 0$ be given. Since $f$ is a bounded, Lebesgue measurable function defined on a Lebesgue measurable set, by The Simple Function Approximation Lemma there exists simple functions $\varphi_{\epsilon}$ and $\psi_{\epsilon}$ such that $\varphi_{\epsilon}(x) \leq f(x) \leq \psi_{\epsilon}(x)$ for all $x \in E$ and $\displaystyle{0 \leq \psi_{\epsilon} - \varphi_{\epsilon}(x) < \frac{\epsilon}{m(E)}}$ for all $x \in E$. So by the monotonicity property of the Lebesgue integral of simple functions we have that:
(3)
\begin{align} \quad 0 \leq \int_E (\psi_{\epsilon} - \varphi_{\epsilon}) < \int_E \frac{\epsilon}{m(E)} = \epsilon \end{align}
• And by the linearity property we have that:
(4)
\begin{align} 0 \leq \int_E \psi_{\epsilon} - \int_E \varphi_{\epsilon} < \epsilon \end{align}
• So for all $\epsilon > 0$ we have that:
(5)
\begin{align} \quad \overline{\int_E} f = \inf \left \{ \int_E \psi : \psi \: \mathrm{is \: a \: simple \: function}, \: \psi(x) \geq f(x) \: \mathrm{on} \: E \right \} \leq \int_E \psi_{\epsilon} & \leq \int_E \varphi_{\epsilon} + \epsilon \\ & \leq \sup \left \{ \int_E \varphi + \epsilon : \varphi \: \mathrm{is \: a \: simple \: function}, \: \varphi(x) \leq f(x) \: \mathrm{on \:} E \right \} \\ & \leq \sup \left \{ \int_E \varphi : \varphi \: \mathrm{is \: a \: simple \: function}, \: \varphi(x) \leq f(x) \: \mathrm{on \:} E \right \} + \epsilon \\ & \leq \underline{\int_E} f + \epsilon \end{align}
• Hence:
(6)
• From $(*)$ and $(**)$ we conclude that:
• So $f$ is Lebesgue integrable on $E$. $\blacksquare$