# Upper Functions and Integrals of Upper Functions

Recall from the Another Comparison Theorem for Integrals of Step Functions on General Intervals page that if $(f_n(x))_{n=1}^{\infty}$ is an increasing sequence of step functions that converge to $f$ almost everywhere on $I$ and if $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx}$ converges then for any step function $g$ such that $g(x) \leq f(x)$ almost everywhere on $I$ we have that:

(1)We will now define a special class of functions known as upper functions below:

Definition: A function $f$ defined on the interval $I$ is said to be an Upper Function on $I$ if there exists an increasing sequence of step functions $(f_n(x))_{n=1}^{\infty}$ such that $(f_n(x))_{n=1}^{\infty}$ is increasing and converges to $f$ almost everywhere on $I$ and such that $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx}$ is finite. The sequences $(f_n(x))_{n=1}^{\infty}$ are called Generating Sequences for $f$, and the Set of All Upper Functions on $I$ is denoted $U(I)$. Furthermore, we define the Integral of $f$ on $I$ as $\displaystyle{\int_I f(x) \: dx = \lim_{n \to \infty} \int_I f_n(x) \: dx}$. |

If we define the integral of an upper function $f$ on $I$ as $\displaystyle{\int_I f(x) \: dx = \lim_{n \to \infty} f_n(x) \: dx}$ for some generating sequence $(f_n(x))_{n=1}^{\infty}$ of $f$, we would like to be sure that the value of the integral is independent on the choice of generating sequence of $f$ to make $\displaystyle{\int_I f(x) \: dx}$ well-defined. The next theorem tells us that this is certainly the case.

Theorem 1: Let $f$ be an upper function on $I$, and let $(f_n(x))_{n=1}^{\infty}$ and $(f_n^*(x))_{n=1}^{\infty}$ both be generating sequences for $f$, i.e., $(f_n(x))_{n=1}^{\infty}$ and $(f_n^*(x))_{n=1}^{\infty}$ are both increasing sequences of step functions that converge to $f$ almost everywhere on $I$ and $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx}$ and $\displaystyle{\lim_{n \to \infty} \int_I f_n^*(x) \: dx}$ are both finite. Then $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx = \lim_{n \to \infty} \int_I f_n^*(x) \: dx}$. |

**Proof:**For each $n \in \mathbb{N}$ we have that the function $f_n^*(x)$ is such that $f_n^*(x) \leq f(x)$ almost everywhere on $I$, and so by the theorem referenced at the top of this page, we have that since $(f_n(x))_{n=1}^{\infty}$ is an increasing sequence of functions that converges to $f$ almost everywhere on $I$ that then:

- Taking the limit as $n \to \infty$ shows us that:

- Similarly, for each $n \in \mathbb{N}$ we have that the function $f_m(x)$ is such that $f_m(x) \leq f(x)$ almost everywhere on $I$ and so by the theorem referenced at the top of this page, we have that since $(f_m(x))_{m=1}^{\infty}$ is an increasing sequence of functions that converges to $f$ almost everywhere on $I$ that then:

- Taking the limit as $n \to \infty$ shows us that:

- Combining $(*)$ and $(**)$ shows us that: