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}