# The Limit of the Integral of a Decreasing Sequence of Nonnegative Step Functions Approaching 0 a.e. on General Intervals

Before we look at the main theorem presented on this page, we will first need to define what it means for a sequence of functions to be increasing or decreasing.

Definition: A sequence of functions $(f_n(x))_{n=1}^{\infty}$ with common domain $X$ is said to be an Increasing sequence of functions if $f_n(x) \leq f_{n+1}(x)$ for all $n \in \mathbb{N}$ and for all $x \in X$. Similarly, the sequence is said to be a Decreasing sequence of functions if $f_n(x) \geq f_{n+1}(x)$ for all $n \in \mathbb{N}$ and for all $x \in X$. Furthermore, that sequence of functions is said to be Monotonic if it is either increasing or decreasing. |

*It is extremely important to note that there is a large difference between saying that $(f_n(x))_{n=1}^{\infty}$ is an "increasing/decreasing sequence of functions" to saying that $(f_n(x))_{n=1}^{\infty}$ is a "sequence of increasing/decreasing functions". The first statement is correct with regards to the definition above - the second is different.*

For example, the following image illustrates part of a decreasing sequence of step functions $(s_n(x))_{n=1}^{\infty}$ that is actually converging to the zero function. Notice that the functions in the sequence need not be decreasing functions though:

Now suppose that we have a decreasing sequence of nonnegative step functions $(s_n(x))_{n=1}^{\infty}$ that approach the zero function almost everywhere on an interval $I$, as pictured above. Intuitively, the value of $\int_I s_n(x) \: dx$ at each $n \in \mathbb{N}$ should be decreasing and should be nonnegative since each $s_n(x)$ is a nonnegative function. Since this sequence of functions approaches $0$ almost everywhere, it seems intuitively clear that $\int_I s_n(x) \: dx$ should approach $0$ as $n \to \infty$. The following theorem assures us that this is true.

Theorem 1: Let $(s_n(x))_{n=1}^{\infty}$ be a decreasing sequence of nonnegative step functions that converge to $0$ almost everywhere on the interval $I$. Then $\displaystyle{\lim_{n \to \infty} \int_I s_n(x) \: dx = 0}$. |

**Proof:**Let $\epsilon > 0$ be given and let $M > 0$ be an upper bound to $s_1(x)$. Then $M$ is an upper bound to $s_n(x)$ for all $n \in \mathbb{N}$ since $(s_n(x))_{n=1}^{\infty}$ is a decreasing sequence of step functions.

- Since $s_1(x)$ is a step function on the interval $I$ there exists a closed and bounded interval $[a, b] \subseteq I$ such that $s_1(x) = 0$ for all $x \in I \setminus [a, b]$. Since $(s_n(x))_{n=1}^{\infty}$ is a decreasing sequence of nonnegative functions, this implies that for all $n \in \mathbb{N}$, $s_n(x) = 0$ for all $x \in I \setminus [a, b]$.

- So in fact, $s_n(x)$ is a step function in the usual sense on $[a, b]$. So, for each $n \in \mathbb{N}$ there exists a partition $P = \{a = x_0, x_1, ..., x_m = b \} \in \mathscr{P}[a, b]$ ($m$ dependent on $n$) such that $s_n(x)$ is constant on each open subinterval $(x_{k-1}, x_k)$ for $k \in \{1, 2, ..., m \}$. Let $D_n$ be the set of endpoints of these intervals, i.e., $D_n = \{ x_0, x_1, ..., x_m \}$. Then each $D_n$ is a finite set, and furthermore if $\displaystyle{D = \bigcup_{n=1}^{\infty} D_n}$ then $D$ is a countable set since $D$ is a countable union of finite sets. From The Measure of Countable Subsets of Real Numbers page we know that then $m(D) = 0$

- Now since $(s_n(x))_{n=1}^{\infty}$ converges to $0$ almost everywhere on $I$ and since for every $n \in \mathbb{N}$, $s_n(x) = 0$ for all $x \in I \setminus [a, b]$ we have that $(s_n(x))_{n=1}^{\infty}$ converges to $0$ almost everywhere on $[a, b]$. So there exists a subset $E \subseteq [a, b]$ with $m(E) = 0$ and such that $(s_n(x))_{n=1}^{\infty}$ converges for all $x \in [a, b] \setminus E$.

- Now let $x \in [a, b] \setminus (E \cup F)$. Then $x \not \in D$ and $x \not \in E$.

- Since $x \not \in D$ so $x$ must be contained in some open interval $B(x) = (c, d) \subset [a, b]$ (and is hence not open of the endpoints of the subdivisions of all step functions $s_n(x)$.

- Since $x \not \in E$ we have that $(s_n(x))_{n=1}^{\infty}$ converges to $0$. For this fixed $x$, the sequence $(s_n(x))_{n=1}^{\infty}$ is a numerical sequence which converges to $0$ and so for $\epsilon_1 = \frac{\epsilon}{b - a + M} > 0$ there exists an $N = N(x) \in \mathbb{N}$ (dependent on $x$) such that if $n \geq N$ then $s_n(x) < \epsilon_1$. We can further say that for all $n \geq N$ and for all $y \in (c, d)$ we have that:

- So the collection $\{ B(x) = (c_x, d_x) : x \in [a, b] \setminus (D \cup E) \}$ is an open covering of $[a, b] \setminus (D \cup E)$. Furthermore, since $m(D) = m(E) = 0$, we see that $m(D \cup E) = 0$ and so there exists an open interval covering $\{ I_k = (e_k, f_k) : k \in K \}$ (where $K$ is a countable indexing set) of $D \cup E$ that covers $D \cup E$, i.e., $\displaystyle{D \cup E \subseteq \bigcup_{k \in K} I_k}$, and such that $\displaystyle{\sum_{k \in K} (f_k - e_k) < \epsilon_1}$. So actually, $\{ B(x) = (c_x, d_x) : x \in [a, b] \setminus (D \cup E) \} \cup \{ I_k = (e_k, f_k) : k \in K \}$ is an open interval covering of $[a, b]$, i.e.:

- But $[a, b]$ is a compact subset of $\mathbb{R}$ and so there exists a finite open subcovering, say:

- Now of the set $\{ B(x_1), B(x_2), ..., B(x_p) \}$ let $N = \max \{ N(x_1), N(x_2), ..., N(x_p) \}$ as chosen in $(*)$. Then we have that for all $n \geq N$ and for all $y \in \bigcup_{k=1}^{p} (c_{x_k}, d_{x_k})$ we have that:

- So let $B = \bigcup_{k=1}^{q} (e_k, f_k)$ and let $A = [a, b] \setminus B$. Then:

- Since $A \subseteq \bigcup_{k=1}^{p} (c_{x_k}, d_{x_k})$ we see that for all $n \geq N$ and for all $y \in A$ we have that:

- Now since $M$ is an upper bound for $s_1(x)$ on $[a, b]$ and since $(s_n(x))_{n=1}^{\infty}$ is a decreasing sequence of functions, we also see that:

- So for all $\epsilon_1 > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ we have that:

- Therefore $\displaystyle{\lim_{n \to \infty} \int_I s_n(x) \: dx = 0}$. $\blacksquare$