Bounded Functions

We now extend our definition of the integral from simple functions to bounded functions supported on a set of finite measure.

Bounded functions with finite support

Approximation with simple functions

Recall the Riemann integral strategy: approximate the area from below and above using rectangles (step functions). We do the same here using simple functions.

Consider the following two values:

I_*(f) = \sup_{\varphi \le f} \int \varphi \, d\mu \quad \text{and} \quad I^*(f) = \inf_{\psi \ge f} \int \psi \, d\mu

where the supremum and infimum are taken over all simple functions $\varphi, \psi$ which vanish on $E^c$ .

Inequality 1: Since $\varphi \le f \le \psi$ , we have $\varphi \le \psi$ everywhere. Since both are simple functions, by property 4 of simple integrals:

\int \varphi \, d\mu \le \int \psi \, d\mu

Taking the supremum over $\varphi$ and infimum over $\psi$ , we get:

\sup_{\varphi \le f} \int \varphi \, d\mu \le \inf_{\psi \ge f} \int \psi \, d\mu \quad \implies \quad I_*(f) \le I^*(f) \quad \textcolor{gray}{\text{--- (1)}}

Closing the gap

To define the integral, we want these two values to be equal. We can show this by constructing a specific sequence of simple functions that converges to $f$ .

Since $f$ is bounded, let $|f(x)| \le M$ for all $x$ . For a fixed integer $n$ , partition the range $[-M, M]$ into intervals of size $M/n$ . Define the sets:

E_k = \left\{ x \in E : \frac{kM}{n} \ge f(x) > \frac{(k-1)M}{n} \right\} \quad \text{for } k = -n, \dots, n \text{ (and } k \ne 0 \text{ logic handles } 0)

Now define the approximating simple functions:

\psi_n(x) = \sum_{k=-n}^{n} \frac{kM}{n} \mathbb{1}_{E_k} \quad \text{(Upper approximation)}

\varphi_n(x) = \sum_{k=-n}^{n} \frac{(k-1)M}{n} \mathbb{1}_{E_k} \quad \text{(Lower approximation)}

We approximate $f$ by rounding its values up ( $\psi_n$ ) or down ( $\varphi_n$ ) to the nearest grid level $kM/n$ . Approximating a bounded function

Note that for any $x \in E$ , the difference between the upper and lower approximation is exactly the step size:

\psi_n(x) - \varphi_n(x) = \frac{M}{n} \mathbb{1}_E(x)

Integrating this difference:

\int (\psi_n - \varphi_n) \, d\mu = \int \frac{M}{n} \mathbb{1}_E \, d\mu = \frac{M}{n} \mu(E)

Now, observe that:

\sup_{\varphi \le f} \int \varphi \, d\mu \ge \int \varphi_n \, d\mu = \int \psi_n \, d\mu - \frac{M}{n} \mu(E) \ge \inf_{\psi \ge f} \int \psi \, d\mu - \frac{M}{n} \mu(E)

Taking the limit as $n \to \infty$ (since $\mu(E) < \infty$ , the term $\frac{M}{n}\mu(E) \to 0$ ):

\sup \int \varphi \ge \inf \int \psi \quad \textcolor{gray}{\text{--- (2)}}

Definition of the integral

Combining (1) and (2), we have equality!

\sup_{\varphi \le f} \int \varphi \, d\mu = \inf_{\psi \ge f} \int \psi \, d\mu

We define the Lebesgue Integral of $f$ to be this common value:

\int f \, d\mu := \sup_{\varphi \le f} \int \varphi \, d\mu