A Comp. Theorem for Integrals of Upper Functions on General Intervals
A Comparison Theorem for Integrals of Upper Functions on General Intervals
Recall from the Upper Functions and Integrals of Upper Functions page that a function $f$ on $I$ 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.
On the Partial Linearity of Integrals of Upper Functions on General Interval page we saw that if $f$ and $g$ were both upper functions on $I$ then $f + g$ is an upper function on $I$ and:
(1)\begin{align} \quad \int_I [f(x) + g(x)] \: dx = \int_I f(x) \: dx + \int_I g(x) \: dx \end{align}
Furthermore, we saw that if $c \in \mathbb{R}$, $c \geq 0$, then $cf$ is an upper function on $I$ and:
(2)\begin{align} \quad \int_I cf(x) \: dx = c \int_I f(x) \: dx \end{align}
We will now look at some more nice properties of integrals of upper functions on general intervals.
| Theorem 1: Let $f$ and $g$ be upper functions on the interval $I$. If $f(x) \leq g(x)$ almost everywhere on $I$ then $\displaystyle{\int_I f(x) \: dx \leq \int_I g(x) \: dx}$ |
- Proof: Let $f$ and $g$ be upper functions on the interval $I$. 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.
- Now since $f(x) \leq g(x)$ and since $(f_n(x))_{n=1}^{\infty}$ is an increasing sequence of step functions that generates $f$ we have that for all $n \in \mathbb{N}$ that:
\begin{align} \quad f_n(x) \leq f(x) \leq g(x) \end{align}
- By applying the theorem presented on the Another Comparison Theorem for Integrals of Step Functions on General Intervals page, we see that then:
\begin{align} \quad \int_I f_n(x) \: dx \leq \lim_{n \to \infty} \int_I g_n(x) \: dx = \int_I g(x) \: dx \end{align}
- Taking $n \to \infty$ gives us that:
\begin{align} \quad \lim_{n \to \infty} \int_I f_n(x) \: dx & \leq \lim_{n \to \infty} \int_I g(x) \: dx \\ \quad \int_I f(x) \: dx & \leq \int_I g(x) \: dx \quad \blacksquare \end{align}
| Corollary 1: Let $f$ and $g$ be upper functions on the interval $I$. If $f(x) = g(x)$ almost everywhere on $I$ then $\displaystyle{\int_I f(x) \: dx = \int_I g(x) \: dx}$ |
- Proof: If $f(x) = g(x)$ almost everywhere on $I$ then $f(x) \leq g(x)$ almost everywhere on $I$ and by Theorem 1 we see that:
\begin{align} \quad \int_I f(x) \: dx \leq \int_I g(x) \: dx \quad (*) \end{align}
- But also $f(x) \geq g(x)$ almost everywhere on $I$ and using Theorem 1 again we have:
\begin{align} \quad \int_I f(x) \: dx \geq \int_I g(x) \: dx \quad (**) \end{align}
- Combining $(*)$ and $(**)$ shows us that $\displaystyle{\int_I f(x) \: dx = \int_I g(x) \: dx}$ as desired. $\blacksquare$