Additivity of Integrals of Upper Functs. on Subintervals of Gen. Intervals

Additivity of Integrals of Upper Functions on Subintervals of General Intervals

We will now look at some nice additivity properties of integrals of upper functions on subintervals of a general interval $I$.

Theorem 1: Let $f$ be an upper function on $I$ and let $I_1, I_2 \subset I$ be subintervals of $I$ such that $I_1$ and $I_2$ intersect in at most one point and such that $I = I_1 \cup I_2$. Then if $f(x) \geq 0$ almost everywhere on $I$ then $f$ is an upper function on both $I_1$ and $I_2$ and furthermore, $\displaystyle{\int_I f(x) \: dx = \int_{I_1} f(x) \: dx + \int_{I_2} f(x) \: dx}$.

• Proof: Let $f$ be an upper function on $I$. Then there exists an increasing sequence of step 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. Define a new sequence of functions $(f_n^+(x))_{n=1}^{\infty}$ where

(1)

\begin{align} \quad f_n^+(x) = \max \{ f_n(x), 0 \} \end{align}

• Since $f(x) \geq 0$ we have that then $(f_n^+(x))$ is a nonnegative increasing sequence of step functions that converges to $f$ almost everywhere on $I$.

• Consider the subinterval $I_1 \subseteq I$. Then we have that $(f_n(x))_{n=1}^{\infty}$ converges to $f$ almost everywhere on $I_1$ and also $(f_n^+(x))_{n=1}^{\infty}$ converges to $f$ almost everywhere on $I_2$. The same goes for the subinterval $I_2 \subseteq I$. So $f$ is an upper function on the subintervals $I_1$ and $I_2$. Furthermore, for each $n \in \mathbb{N}$ we have that:

(2)

\begin{align} \quad \int_I f_n^+(x) \: dx = \int_{I_1} f_n^+(x) \: dx + \int_{I_2} f_n^+n(x) \: dx \end{align}

• Taking the limit as $n \to \infty$ and we get that:

(3)

\begin{align} \quad \lim_{n \to \infty} \int_I f_n^+(x) \: dx &= \lim_{n \to \infty} \int_{I_1} f_n^+(x) \: dx + \lim_{n \to \infty} \int_{I_2} f_n^+(x) \: dx \\ \quad \int_I f(x) \: dx &= \int_{I_1} f(x) \: dx + \int_{I_2} f(x) \: dx \quad \blacksquare \end{align}

Theorem 2: Let $I$ be an interval and let $I_1, I_2 \subset I$ be intervals such that $I_1$ and $I_2$ intersects in at most one point and such that $I = I_1 \cup I_2$. If $g$ is an upper function defined on $I_1$ and $h$ is an upper function defined on $I_2$ then the function $f(x) = \left\{\begin{matrix} g(x) & x \in I_1\\ h(x) & x \in I_2 \setminus \{I_1 \cap I_2\} \end{matrix}\right.$ is an upper function on the entire $I$ and $\displaystyle{\int_I f(x) \: dx = \int_{I_1} g(x) \: dx + \int_{I_2} h(x) \: dx}$.

• Proof: Let $g$ and $h$ be upper functions defined on $I_1$ and $I_2$ respectively. Then there exists increasing sequences of functions $(g_n(x))_{n=1}^{\infty}$ and $(h_n(x))_{n=1}^{\infty}$ that converge to $g$ and $h$ (respectively) almost everywhere on $I_1$ and $I_2$ (respectively) and such that $\displaystyle{\lim_{n \to \infty} \int_{I_1} g_n(x) \: dx}$ and $\displaystyle{\lim_{n \to \infty} \int_{I_2} h_n(x) \: dx}$ are finite.

• For $x \in I \setminus (I_1 \cap I_2)$ define $g_n(x) = 0$, and for $x \in I_1$ define $h_n(x) = 0$. Then the sequence of functions $(g_n(x) + h_n(x))_{n=1}^{\infty}$ is increasing and coverges to $g + h$ almost everywhere on the entire interval $I$. Furthermore, $\lim_{n \to \infty} \int_I [g_n(x) + h_n(x)] \: dx$ is finite, and so we see that:

(4)

\begin{align} \quad \lim_{n \to \infty} \int_I [g_n(x) + h_n(x)] \: dx &= \lim_{n \to \infty} \int_I g_n(x) \: dx + \lim_{n \to \infty} \int_I h_n(x) \\ \quad &= \lim_{n \to \infty} \int_{I_1} g_n(x) \: dx + \lim_{n \to \infty} \int_{I_2 \setminus (I_1 \cap I_2)} g_n(x) \: dx + \lim_{n \to \infty} \int_{I_2 \setminus (I_1 \cap I_2)} h_n(x) \: dx + \lim_{n \to \infty} \int_{I_1} h_n(x) \: dx \\ \quad &= \lim_{n \to \infty} \int_{I_1} g_n(x) \: dx + \lim_{n \to \infty} \int_{I_2 \setminus (I_1 \cap I_2)} 0 \: dx + \lim_{n \to \infty} \int_{I_2 \setminus (I_1 \cap I_2)} h_n(x) \: dx + \lim_{n \to \infty} \int_{I_1} 0 \: dx \\ \quad &= \lim_{n \to \infty} \int_{I_1} g_n(x) \: dx + \lim_{n \to \infty} \int_{I_2 \setminus (I_1 \cap I_2)} h_n(x) \: dx \\ \quad &= \int_{I_1} g(x) \: dx + \int_{I_2} h(x) \: dx \quad \blacksquare \end{align}