Cardinality of the set of bijective Lebesgue-measurable functions on the unit intervall Let $I= [0,1]$ be the unit intervall. It is well known that the cardinality of $$B = \{ f: I \to I \mid f \text{ bijective} \}$$ and $$L = \{ f: I \to I \mid f \text{ Lebesgue-measurable} \}$$ is $2^{|\mathbb{R}|}$. 

Question: What is the cardinality of $B \cap L$ ?
The cardinality of the set all bijective, Lebesgue-measurable functions $f: I \to I$ is $2^{|\mathbb{R}|}$. 


Proof: Wlog we can take $I = [-1, 1]$. 

The idea is to associate each measurable subset $M \subseteq (0,1]$ a measurable bijection $f_M$. The construction of $f_M$ follows the accepted answer of Cardinality of real bijective functions/injective functions from $\mathbb{R}$ to $\mathbb{R}$.  

Let $\chi_M: [0,1] \to I$ be the characteristic function of $M$

\begin{equation}
    \chi_M(x)=
    \begin{cases}
      1, &amp; \text{if}\ x \in M \\
      -1, &amp; \text{if}\ x \not\in M
    \end{cases}
  \end{equation}
and let $$f_M: I \to I, \, x \mapsto \chi_M(|x|)\cdot x.$$ 
$f_M$ satisfies $f_M \circ f_M = \text{id}_I$ and is thus a bijection. Moreover, $f_M$ can be represented by
$$f_M = (\chi_M \circ \text{abs}) \cdot \text{id}_M$$
as a composition and product of measurable functions (note: the characteristic function of a measurable set is measurable) and is therefore measurable. 

The cardinality of measurable subsets of $(0,1]$ is $2^{|\mathbb{R}|}$. In fact, the Cantor set $C$ has cardinality $|\mathbb{R}|$ and has measure $0$. Hence all of the $2^{|\mathbb{R}|}$ subsets of $C$ are also measurable (with measure $0$). Since the assignment $M \to f_M$ is injective, we have found $2^{|\mathbb{R}|}$ bijective, measurable functions $f_M$ and since the cardinality of $I^I$ is $2^{|\mathbb{R}|}$, there are  $2^{|\mathbb{R}|}$ bijective, measurable functions $I \to I$. q.e.d. 

Remark: It's easy to see that $f_M$ is non-measurable, if $M$ is non-measurable. Since there are $2^{|\mathbb{R}|}$ non-measurable subsets of $I$ (in ZFC), the same proof shows: 


  The cardinality of the set of all bijective, non-Lebesgue-measurable functions $f: I \to I$ is $2^{|\mathbb{R}|}$.
Let $C$ be the Cantor set, or any other Lebesgue-null set of cardinality $\mathfrak{c}$  (I'll write $\mathfrak{c}$ instead of $|\mathbb{R}|$).  Let $g : C \to C$ be any bijection; there are $2^\mathfrak{c}$ of them.  For each $g$, consider the function $f_g : I \to I$ defined by
$$f_g(x) = \begin{cases} g(x), &amp; x \in C \\ x, &amp; x \notin C.\end{cases}$$
Then each $f_g$ is clearly a bijection, and $f_g = f_h$ iff $g=h$.  Moreover $f_g$ is Lebesgue measurable because it is a.e. equal to the measurable function $f(x) = x$.  So there are at least $2^\mathfrak{c}$ measurable bijections, and you already know there are at most that many.