# 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.