On cardinality of equivalence classes 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.
 A: 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.
