Upper Functions and Integrals of Upper Functions

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)
\begin{align} \quad \int_I g(x) \: dx \leq \lim_{n \to \infty} \int_I f_n(x) \: dx \end{align}

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:
(2)
\begin{align} \quad \int_I f_n^*(x) \: dx \leq \lim_{m \to \infty} \int_I f_m(x) \: dx \end{align}
• Taking the limit as $n \to \infty$ shows us that:
(3)
\begin{align} \quad \lim_{n \to \infty} \int_I f_n^*(x) \: dx \leq \lim_{m \to \infty} \int_I f_m(x) \: dx \\ \quad \lim_{n \to \infty} \int_I f_n^*(x) \: dx \leq \lim_{n \to \infty} \int_I f_n(x) \: dx \quad (*) \end{align}
• 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:
(4)
\begin{align} \quad \int_I f_n(x) \: dx \leq \lim_{m \to \infty} \int_I f_m^*(x) \: dx \end{align}
• Taking the limit as $n \to \infty$ shows us that:
(5)
\begin{align} \quad \lim_{n \to \infty} \int_I f_n(x) \: dx \leq \lim_{m \to \infty} \int_I f_m^*(x) \: dx \\ \quad \lim_{n \to \infty} \int_I f_n(x) \: dx \leq \lim_{n \to \infty} \int_I f_n^*(x) \: dx \quad (**) \end{align}
• Combining $(*)$ and $(**)$ shows us that:
(6)
\begin{align} \quad \lim_{n \to \infty} \int_I f_n(x) \: dx = \lim_{n \to \infty} \int_I f_n^*(x) \: dx \quad \blacksquare \end{align}