What is the cardinality of the set $\mathcal{R}[0,1]$ of all Riemann integrable real functions on [0,1]?
I expect it to be $2^\mathfrak{c}$. A function is Riemann integrable if and only if it is continuous almost everywhere and bounded. Since a set of measure 0 can be uncountable, I assume one can construct $2^\mathfrak{c}$ subsets of $[0,1]$ that have measure 0 yet are discontinuity sets of real functions. But then I also realize that there can be measure zero sets which are never discontinuity sets of a real function. So, I am stuck at this point and have no idea how to proceed.
Your guess is correct. The trick is to not worry about what the exact set of discontinuity is but instead just find a large family of sets which can only be discontinuous on some uncountable set of measure zero. For instance, if $C$ is the Cantor set and $A\subseteq C$ is any subset, the characteristic function of $A$ is Riemann integrable (since it is continuous at least on all of $[0,1]\setminus C$). There are $2^{\mathfrak{c}}$ such subsets, and so there are $2^{\mathfrak{c}}$ Riemann integrable functions.