Another Comp. Theorem for Integrals of Step Functs. on Gen. Intervals

Another Comparison Theorem for Integrals of Step Functions on General Intervals

Recall from the The Limit of the Integral of a Decreasing Sequence of Nonnegative Step Functions Approaching 0 a.e. on General Intervals page that if $(s_n(x))_{n=1}^{\infty}$ is a decreasing sequence of nonnegative step functions that converge to $0$ almost everywhere on the interval $I$ then:

(1)
\begin{align} \quad \lim_{n \to \infty} \int_I s_n(x) \: dx = 0 \end{align}

We will now use that very important result to prove a nice comparison theorem for integrals of step functions on general intervals.

 Theorem 1: Let $(f_n(x))_{n=1}^{\infty}$ be an increasing sequence of step functions that converge to $f$ almost everywhere on $I$, and suppose that $\displaystyle{\lim_{n \to \infty} \int_I f_n(x) \: dx}$ exists. Then for any step function $g$ such that $g(x) \leq f(x)$ almost everywhere on $I$ we have that $\displaystyle{\int_I g(x) \: dx \leq \lim_{n \to \infty} \int_I f_n(x) \: dx}$.
• Proof: Define a new sequence of functions $(s_n(x))_{n=1}^{\infty}$ for each $n \in \mathbb{N}$ by:
(2)
\begin{align} \quad s_n(x) = \max \{ g(x) - f_n(x), 0 \} \end{align}
• Then $(s_n(x))_{n=1}^{\infty}$ is a decreasing sequence of functions (since $g(x)$ is a fixed function and $(f_n(x))_{n=1}^{\infty}$ is an increasing sequence of functions, so $(-f_n(x))_{n=1}^{\infty}$ is a decreasing sequence of functions). Furthermore, $(s_n(x))_{n=1}^{\infty}$ is nonnegative, and $(s_n(x))_{n=1}^{\infty}$ converges to $0$ almost everywhere on $I$. Therefore we have that:
(3)
\begin{align} \quad \lim_{n \to \infty} \int_I s_n(x) \: dx = 0 \quad (*) \end{align}
• So for each $n \in \mathbb{N}$ we have that:
(4)
\begin{align} \quad \int_I s_n(x) \: dx \leq \int_I [g(x) - f_n(x)] \: dx = \int_I g(x) \: dx - \int_I f_n(x) \: dx \end{align}
• Taking the limit as $n \to \infty$ of both sides and using $(*)$ yields:
(5)
\begin{align} \quad \lim_{n \to \infty} \int_I s_n(x) \: dx & \geq \lim_{n \to \infty} \int_I g(x) \: dx - \lim_{n \to \infty} \int_I f_n(x) \: dx \\ \quad 0 & \geq \int_I g(x) \: dx - \lim_{n \to \infty} \int_I f_n(x) \: dx \\ \quad \lim_{n \to \infty} \int_I f_n(x) \: dx & \geq \int_I g(x) \: dx \end{align}
• So $\displaystyle{\int_I g(x) \: dx \leq \lim_{n \to \infty} \int_I f_n(x) \: dx}$. $\blacksquare$