Partial Linearity of Integrals of Upper Functions on General Intervals

Recall from the Upper Functions and Integrals of Upper Functions page that a function $f$ defined on an interval $I$ is said to be an upper function on $I$ if there exists an increasing sequence of functions $(f_n(x))_{n=1}^{\infty}$ that converges to $f$ almost everywhere on $I$ and such that $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx}$ is finite.

Furthermore, we defined the integral of $f$ on $I$ to be:

(1)

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

We also noted that the choice of generating sequence $(f_n(x))_{n=1}^{\infty}$ of $f$ did not matter as long as it satisfied the conditions above.

In the following theorems we will see that the part of the linearity properties of integrals of upper functions hold.

Theorem 1: Let $f$ and $g$ both be upper functions defined on $I$. Then $f + g$ is an upper function on $I$ and $\displaystyle{\int_I [f(x) + g(x)] \: dx = \int_I f(x) \: dx + \int_I g(x) \: dx}$.

• Proof: Let $f$ and $g$ both be upper functions defined on $I$. Then there exists increasing sequences $(f_n(x))_{n=1}^{\infty}$ and $(g_n(x))_{n=1}^{\infty}$ that converge to $f$ and $g$ (respectively) almost everywhere on $I$ and such that $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx}$ and $\displaystyle{\lim_{n \to \infty} \int_I g_n(x) \: dx}$ are finite.

• Consider the sequence $(f_n(x) + g_n(x))_{n=1}^{\infty}$. Then this is an increasing sequence of functions that converges to $f + g$ almost everywhere on $I$ and furthermore, $\displaystyle{\lim_{n \to \infty} \int_I [f_n(x) + g_n(x)] \: dx = \lim_{n \to \infty} \int_I f_n(x) \: dx + \lim_{n \to \infty} \int_I g_n(x) \: dx}$ which is finite. So $f +g$ is an upper function on $I$ and furthermore:

(2)

\begin{align} \quad \int_I [f(x) + g(x)] &= \lim_{n \to \infty} \int_I [f_n(x) + g_n(x)] \: dx \\ \quad &= \lim_{n \to \infty} \int_I f_n(x) \: dx + \lim_{n \to \infty} \int_I g_n(x) \: dx \\ \quad &= \int_I f(x) \: dx + \int_I g(x) \: dx \quad \blacksquare \end{align}

Theorem 2: Let $f$ be an upper function defined on $I$ and let $c \in \mathbb{R}$, $c \geq 0$. Then $cf$ is an upper function on $I$ and $\displaystyle{\int_I cf(x) \: dx = c \int_I f(x) \: dx}$.

It is very important to note that $c \geq 0$.

• Proof: Let $f$ be an upper function defined on $I$. Then there exists an increasing sequence $(f_n(x))_{n=1}^{\infty}$ that converges to $f$ almost everywhere on $I$ and such that $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx}$ is finite.

• For $c \geq 0$, consider the sequence $(cf_n(x))_{n=1}^{\infty}$. Then this is an increasing sequence of functions that converges to $cf$ almost everywhere on $I$ and furthermore, $\displaystyle{\lim_{n \to \infty} \int_ cf_n(x) \: dx = c \lim_{n \to \infty} \int_I f_n(x) \: dx}$ is finite. So $cf$ is an upper function on $I$ and furthermore:

(3)

\begin{align} \quad \int_I cf(x) \: dx &= \lim_{n \to \infty} \int_I cf_n(x) \: dx \\ \quad &= c \lim_{n \to \infty} \int_I f_n(x) \: dx \\ \quad &= c \int_I f(x) \: dx \quad \blacksquare \end{align}