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:

\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:

\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:

\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:

\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 $\displaystyle{\int_I g(x) \: dx \leq \lim_{n \to \infty} \int_I f_n(x) \: dx}$. $\blacksquare$