# The Unapologetic Mathematician

## Simple and Elementary Functions

We now introduce two classes of functions that are very easy to work with. As usual, we’re working in some measurable space $(X,\mathcal{S})$.

First, we have the “simple functions”. Such a function is described by picking a finite number of pairwise disjoint measurable sets $\{E_i\}_{i=1}^n\subseteq\mathcal{S}$ and a corresponding set of finite real numbers $\alpha_i$. We use these to define a function by declaring $f(x)=\alpha_i$ if $x\in E_i$, and $f(x)=0$ if $x$ is in none of the $E_i$. The very simplet example is the characteristic function $\chi_E$ of a measurable function $E$. Any other simple function can be written as $\displaystyle f(x)=\sum\limits_{i=1}^n\alpha_i\chi_{E_i}(x)$

Any simple function is measurable, for the preimage $f^{-1}(A)$ is the union of all the $E_i$ corresponding to those $\alpha_i\in A$, and is thus measurable.

It’s straightforward to verify that the product and sum of any two simple functions is itself a simple function — given functions $f=\sum\alpha_i\chi_{E_i}$ and $g=\sum\beta_j\chi_{F_j}$, we have $fg=\sum\alpha_i\beta_j\chi_{E_i\cap F_j}$ and $f+g=\sum(\alpha_i+\beta_j)\chi_{E_i\cap F_j}$. It’s even easier to see that any scalar multiple of a simple function is simple — $cf=\sum c\alpha_i\chi_{E_i}$. And thus the collection of simple functions forms a subalgebra of the algebra of measurable functions.

“Elementary functions” are similar to simple functions. We slightly relax the conditions by allowing a countably infinite number of measurable sets $E_i$ and corresponding values $\alpha_i$.

Now, why do we care about simple functions? As it happens, every measurable function can be approximated by simple functions! That is, given any measurable function $f$ we can find a sequence $f_n$ of simple functions converging pointwise to $f$.

To see this, first break $f$ up into its positive and negative parts $f^+$ and $f^-$. If we can approximate any nonnegative measurable function by a pointwise-increasing sequence of nonnegative simple functions, then we can approximate each of $f^+$ and $f^-$, and the difference of these series approximates $f$. So, without loss of generality, we will assume that $f$ is nonnegative.

Okay, so here’s how we’ll define the simple functions $f_n$: \displaystyle f_n(x)=\left\{\begin{aligned}\frac{i-1}{2^n}\qquad&\frac{i-1}{2^n}\leq f(x)<\frac{i}{2^n},\quad i=1,\dots,n2^n\\n\qquad&n\leq f(x)\end{aligned}\right.

That is, to define $f_n$ we chop up the nonnegative real numbers $\left[0,n\right)$ into $n2^n$ chunks of width $2^n$, and within each of these slices we round values of $f$ down to the lower endpoint. If $f(x)\geq n$, we round all the way down to $n$. There can only ever be $n2^n+1$ values for $f_n$, and each of these corresponds to a measurable set. The value $\frac{i-1}{2^n}$ corresponds to the set $\displaystyle f^{-1}\left(\left[\frac{i-1}{2^n},\frac{i}{2^n}\right)\right)$

while the value $n$ corresponds to the set $f^{-1}\left(\left[n,\infty\right]\right)$. And thus $f_n$ is indeed a simple function.

So, does the sequence $\{f_n\}$ converge pointwise to $f$? Well, if $f(x)=\infty$, then $f_n(x)=n$ for all $n$. On the other hand, if $k\leq f(x) then $f_k(x)=k$; after this point, $f_n(x)$ and $f(x)$ are both within a slice of width $\frac{1}{2^n}$, and so $0\leq f(x)-f_n(x)<\frac{1}{2^n}$. And so given a large enough $n$ we can bring $f_n(x)$ within any desired bound of $f(x)$. Thus the sequence $\{f_n\}$ increases pointwise to the function $f$.

But that’s not all! If $f$ is bounded above by some integer $N$, the sequence $f_n$ converges uniformly to $f$. Indeed, once we get to $n\geq N$, we cannot have $f_n(x)=n$ for any $x\in X$. That is, for sufficiently large $n$ we always have $0\leq f(x)-f_n(x)<\frac{1}{2^n}$. Given an $\epsilon>0$ we pick an $n$ so that both $n\geq N$ and $\frac{1}{2^n}<\epsilon$, and this $n$ will guarantee $\lvert f(x)-f_n(x)\rvert<\epsilon$ for every $x\in X$. That is: the convergence is uniform.

This is also where elementary functions come in handy. If we’re allowed to use a countably infinite number of values, we can get uniform convergence without having to ask that $f$ be bounded. Indeed, instead of defining $f_n(x)=n$ for $f(x)\geq n$, just chop up all positive values into slices of width $\frac{1}{2^n}$. There are only a countably infinite number of such slices, and so the resulting function $f_n$ is elementary, if not quite simple.

May 11, 2010 Posted by | Analysis, Measure Theory | 12 Comments