The Unapologetic Mathematician

Mathematics for the interested outsider

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)<k+1 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


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