The Lebesgue Integral of Bounded Functions

So far we have defined the Lebesgue integral of a simple function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$.

We are now ready to define the Lebesgue integral of a bounded function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. We first define the upper and lower Lebesgue integrals of such a function $f$.

Definition: Let $f$ be a bounded function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. The Upper Lebesgue Integral of $f$ on $E$ is $\displaystyle{(L) \overline{\int_E} f = \inf \left \{ \int_E \psi : \psi \: \mathrm{is \: a \: simple \: function}, \: \psi(x) \geq f(x) \: \mathrm{on \:} E \right \}}$ and the Lower Lebesgue Integral of $f$ on $E$ is $\displaystyle{(L) \underline{\int_E} f = \sup \left \{ \int_E \varphi : \varphi \: \mathrm{is \: a \: simple \: function}, \: \varphi(x) \leq f(x) \: \mathrm{on \:} E \right \}}$.

We now define the Lebesgue integral for $f$.

Definition: Let $f$ be a bounded function defined on a Lebesgue measurable set $E$ with $m(E) < \infty$. $f$ is said to be Lebesgue Integrable on $E$ if $\displaystyle{(L) \overline{\int_E} f = (L) \underline{\int_E} f}$.

For example, consider the function $f(x) = x$ defined on $E = [0, 1]$. For each $n \in \mathbb{N}$, consider the sequence of step functions:

(1)

\begin{align} \quad \varphi_n(x) = \left\{\begin{matrix} 0 & \mathrm{if} \: x \in \left [0, \frac{1}{n} \right ]\\ \frac{1}{n} & \mathrm{if} \: x \in \left [ \frac{1}{n}, \frac{2}{n} \right ]\\ \\ \vdots & \\ \frac{n-1}{n} & \mathrm{if} \: x \in \left [\frac{n-1}{n}, 1\right ]\\ \end{matrix}\right. \\ \quad \psi_n(x) = \left\{\begin{matrix} \frac{1}{n} & \mathrm{if} \: x \in \left [0, \frac{1}{n} \right ]\\ \frac{2}{n} & \mathrm{if} \: x \in \left [ \frac{1}{n}, \frac{2}{n} \right ]\\ \\ \vdots & \\ 1 & \mathrm{if} \: x \in \left [\frac{n-1}{n}, 1\right ]\\ \end{matrix}\right. \end{align}

Then for each $n \in \mathbb{N}$ it is easy to see that $\varphi_n(x) \leq f(x) \leq \psi_n(x)$. Furthermore,

(2)

\begin{align} \quad (L) \int_{[0, 1]} \varphi_n(x) = 0 \cdot \frac{1}{n} + \frac{1}{n} \cdot \frac{1}{n} + ... + \frac{n-1}{n} \cdot \frac{1}{n} = \left ( \frac{1}{n} + \frac{2}{n} + ... + \frac{n-1}{n} \right ) \cdot \frac{1}{n} = \frac{\frac{(n-1)n}{2}}{n^2} = \frac{n^2 - n}{2n^2} = \frac{1}{2} - \frac{1}{2n} \\ \quad (L) \int_{[0, 1]} \psi_n(x) = \frac{1}{n} \cdot \frac{1}{n} + \frac{2}{n} \cdot \frac{1}{n} + ... + 1 \cdot \frac{1}{n} = \left ( \frac{1}{n} + \frac{2}{n} + ... + \frac{n}{n} \right ) \cdot \frac{1}{n} = \frac{\frac{n(n+1)}{2}}{n^2} = \frac{n^2 + n}{2n^2} = \frac{1}{2} + \frac{1}{2n} \end{align}

As $n \to \infty$ we see that:

(3)

\begin{align} \quad \frac{1}{2} \leq (L) \underline{\int_{[0,1]}} f \leq (L) \overline{\int_{[0,1]}} f \leq \frac{1}{2} \end{align}

Therefore the upper and lower Lebesgue integrals of $f$ on $E$ are equal and:

(4)

\begin{align} \quad (L) \int_{[0, 1]} f = \frac{1}{2} \end{align}