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Say $R$ is the set of all Riemann integrable functions $f:[a,b]\rightarrow \mathbb{R}$. We define an equivalence relation in $R$ as follows: $f$ and $g$ are said to be integrally equivalent iff they differ only on a (Lebesgue) measure zero set. Prove that the collection of these equivalence classes has a cardinality of a continuum.

Please let me remind that a function is Riemann integrable on $[a,b]$ iff it is bounded there and its discontinuity set there is a null set (has measure zero).

It is known that $R$ has cardinality of $2^{|\mathbb{R}|}$, that is, greater that the continuum. Also, the collection of null sets has the same cardinality.

Please give your opinions on how one could approach this problem. Complete proofs are also welcome. Thank you.

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I just have to run out, I've deleted my answer but when I return I might undelete and correct it (granted no solution similar was posted). – Asaf Karagila Sep 3 '12 at 18:26
up vote 4 down vote accepted

I think you can show that continuous functions are dense in $R/_\sim$ endowed with the integral norm in much the same way as you do for $L^1$ (or just embed into $L^1$ or $L^\infty$ if you're feeling lazy). This shows that $R$ is separable (because continuous functions are, by piecewise linear functions glued at rational points) separable metric space can't have cardinality larger than continuum (because for every point we have a sequence converging to it, and there are only continuum many of those).

For the lower bound, we can of course use constant functions.

A more involved proof: we can show that piecewise constant functions with rational values and rational jump points are dense in $R/_\sim$ with integral norm. We can assume wlog that $a=0,b=1$. Choose arbitrary $f\in R,\varepsilon>0$. When we put

$$S(f,n,j):=\sup_{x\in [(j-1)/n,j/n]}f(x)$$ $$U(f,n):=1/n \cdot \sum_{j=1}^n S(f,n,j)$$ , $I(f,n,j),L(f,n)$ likewise, and choose $N$ large enough so that $U(f,N)-L(f,N)<\varepsilon/2$. For each $j\le N$ we can choose $q_j$ rational so that $\varepsilon/2>S(f,N,j)-q_j>0$. Then put $g(x)=q_j$ for $x\in [(j-1)/N,j/N)$, $g(1)=q_N$. Then:

$$ S(\lvert g-f\rvert,N,j)\le \varepsilon/2+S(f,N,j)-I(f,N,j)$$ so $$U(\lvert g-f\rvert,N)=1/N\cdot \sum_{j=1}^n S(\lvert g-f\rvert,N,j)\le\varepsilon/2+U(f,N)-L(f,N)<\varepsilon$$ And we're done.

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Okay. But I believe you have some fancy theorems behind your proof and I don't really see why it works. Could you please explain your proof in more details? However, I believe there is a more elementary approach to the problem. – Lukas Sep 4 '12 at 10:04
@Lukas: Since you said that Riemann integrable functions are those which are discontinuous on a null set, I assumed that you're mostly comfortable with measure theory. :) You do see that continuous functions with integral norm are separable? – tomasz Sep 4 '12 at 12:44
@Lukas: I added a proof more tailored to your needs. – tomasz Sep 4 '12 at 13:41
A very nice proof indeed. I like that. As I understood, here is how you complete the solution to the original problem: to each class you can assign at least one unique sequence of piecewise constant functions with rational values and jumps on rational points. You have $\mathbb{Q}^{\mathbb{Q}}$,i.e. continuum, of these functions hence a continuum of those sequences hence no more than a continuum of equiv. classes. Thanks. – Lukas Sep 4 '12 at 15:06
@Lukas: actually, there are only countably many such functions. We choose finitely many jump points out of rationals (there are countably many ways to do that), and for each interval we assign a rational number (again, countably many ways to do that). Not that it matters all that much in the end. You can write it as $\bigcup_n {{(\mathbf Q\cap [0,1])}\choose n}\times \mathbf Q^n$ with the obvious identifications. – tomasz Sep 4 '12 at 18:56

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