Riemann Integrability Question

1. Nov 29, 2011

"pi"mp

This isn't a homework question. My adviser has me studying basic analysis and has lately pushed me towards the following question:

"Let f be any continuous function. Can we prove that there exists a SEQUENCE of step functions that converges UNIFORMLY to f?"

I have noticed this idea is important to Riemann integrability but I can't seem to prove this. Or notice why the idea of uniform convergence is so key.

Any help? Thank you

2. Nov 29, 2011

micromass

Functions which are the uniform limit of step functions are usually called "regulated functions". The uniform limit is important as it allows us to define a non-ambiguous integral of such a function. We might not do the same for pointswize limits.

Not all Riemann integrable functions are regulated. Indeed, it is a theorem that the set of discontinuities of a regulated function is countable. It is also known that a function is Riemann integrable iff it's set of discontinuities has Lebesgue measure zero.

Now, the Cantor set C has Lebesgue measure zero, but is not countable. So the function

$$f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow \left\{\begin{array}{c} 1~\text{if}~x\in C\\ 0~\text{if}~x\notin C \end{array}\right.$$

is Riemann integrable but not regulated.

Your question now asks whether a continuous function $f:[a,b]\rightarrow \mathbb{R}$ is regulated. This is indeed the case.
A key point here is that f is uniform continuous. So we can determine for each epsilon a delta such that

$$|x-y|<2\delta~\Rightarrow~|f(x)-f(y)|<\epsilon$$

Now, how do we approximate f by step functions. Well, let $[a,b]=[a,a+\delta]\cup [a+\delta,a+2\delta]\cup ...\cup [a+n\delta,b]$ be a "partition" such that each part has length $<\delta$. Then we define

$$g(x)=g(a+k\delta)~\text{if}~x\in [a+k\delta,a+(k+1)\delta]$$

Then the uniform distance between g and f is smaller than epsilon.

I'll leave the details and verification to you!

Do not that basically all functions we need in our daily life will be regulated.

Last edited: Nov 29, 2011
3. Nov 29, 2011

jgens

Micromass beat me to it :( But to add one small point to his post, it is important to note that the compactness of [a,b] is crucial.

4. Nov 30, 2011

"pi"mp

micromass, why is it important that f itself be uniformly continuous? That wasn't one of the assumptions I was given. Shouldn't this work even if f itself is non-uniformly continuous? Thanks for your reply.

5. Nov 30, 2011

"pi"mp

ah I think I answered my own question. It is because it is defined on a compact set, correct?

6. Nov 30, 2011

micromass

Yes: continuous on a compact set is uniform continuous.