Measurable function that does not map a set of measure zero to a set of measure zero

Question

I'm looking for an example a measurable function that does not map a set of measure zero to a set of measure zero. (we are discussing absolutely continuous functions in class, and the fact that they map sets of measure zero to sets of measure zero)

Answer 1

Consider the standard Cantor set $C$. We know that the Cantor set can be thought of as real numbers in $[0,1]$ expressed in base $3$ with no digit equal to $1$ and it is an easy exercise to prove that it has measure $0$. Anyway, as said before, any element in the Cantor set can be written as

$$x = \sum_{n=1}^{\infty}\frac{a_n}{3^n}, \text{ where } a_n=0,2 \hspace{10px}(\forall n\in \mathbb{N})$$

Define $f: [0,1] \to [0,1]$ as follows:

1. Write the expansion for $x \in [0,1]$ in base $3$.
2. if $x$ contains a $1$ in its expansion, truncate the expansion right after the first instance of $1$.
3. Replace the remaining $2$'s with $1$'s after truncation.
4. Now you have a binary expansion for a real number in $[0,1]$. Convert it back to the decimal system if you want.

Since the elements of the Cantor set have no $1$'s, elements of the Cantor set are never truncated at step II. So, if $0.a_1a_2a_3\cdots$ is a number in $C$, the result of the function will be something like $0.b_1b_2b_3\cdots$ where $b_n = \frac{a_n}{2}$. This means that given a number in $y$ in $[0,1]$, we can first write down its binary expansion $y=0.c_1c_2c_3\cdots$ and then see that $x=(2c_1)(2c_2)(2c_3)\cdots$ is mapped to $y$, i.e. $f(x)=y$. This shows that $f$ maps the Cantor set onto $[0,1]$ and hence, $f$ maps a set with Lebesgue measure $0$ to a set of Lebesgue measure $1$. Q.E.D.

By the way, the function $f$ constructed above is called the Devil's staircase or the Cantor function sometimes and it is an important function for constructing counterexamples in measure theory.

Answer 2

How about the space $X = \{1,2\}$ with the measure $\mu:\mathcal{P}(X) \to \mathbb R$ defined by $\mu(\{1\}) = 0$ and $\mu(\{2\}) = 1$, and the function $f: X \to X$ defined by $f(x) = 2$? The measure $0$ set $\{1\}$ is mapped to the positive-measure set $\{2\}$.