$f:[0, 1]\to \Bbb{R}$ be a function.

  1. $f$ satisfy the property $\mathcal{M}$ of $f(A) $ is meagre for every $A\subset [0, 1]$ meagre.

  2. $f$ satisfy the property $\mathcal{B}$ if $f(A) $ is a set with the property of Baire for every $A\subset [0, 1]$ having the property of Baire.

I don't know whether the property $\mathcal{M}, \mathcal{B}$ has any name or not.

$f\in \Bbb{R}^{[0, 1]}$ with property $\mathcal{B}$ and we know every meager set satisfy the property of Baire this implies image of every meager set is a set with the baire property and this means it is a symmetric difference of an open set and meager set.

I believe we can find $f\in \Bbb{R}^{[0, 1]}$ with the property $\mathcal{B}$ which doesn't satisfy the property $\mathcal{M}$.

Question : Can you give me an explicit example of $f\in \Bbb{R}^{[0, 1]}$ which map every set with Baire property to a set with Baire property but doesn't map a meager set to a meager set?

Question : Can we find $f\in \Bbb{R}^{[0, 1]}$ with the property $\mathcal{M}$ which doesn't satisfy the property $\mathcal{B}$?

The sets with the Baire property forms a $\sigma$-algebra generated by open sets and meagre sets.

Suppose $F\in \Bbb{R}^{[0, 1]}$ satisfy the property $\mathcal{M}$.

$F(M) \subset \Bbb{R}$ is meagre for $\forall M\subset [0, 1]$ meagre. Hence $F$ maps the Cantor set $\mathcal{C}$ and all subsets of $\mathcal{C}$ to meagre set.

So $F(\mathcal{C}) =\mathcal{C}$.

And suppose $F(U) $ is not open for some $U\subset [0, 1]\setminus\mathcal{C}$ open.

Now again we have to map every meagre subset of $U$ to a meagre set. This is difficult and how to map rest of points.

Does this type of function exists? If yes how to construct?


1 Answer 1


Claim 1: Property $\mathcal{B}$ implies property $\mathcal{M}$.

Proof: Suppose $f:[0, 1] \to \mathbb{R}$ is a counterexample and fix a meager $M \subseteq [0, 1]$ such that $f[M]$ is non-meager and has BP (Baire property). Let $W \subseteq f[M]$ be non-meager and without BP (for example, $W = B \cap f[M]$ where $B$ is a Bernstein set). Let $N = M \cap f^{-1}[W]$. Then $N$ is meager and $f[N] = W$ does not have BP. Contradiction.

Claim 2: Property $\mathcal{M}$ does not imply property $\mathcal{B}$.

Proof: Let $V$ be a Vitali set. Define $f:[0, 1] \to \mathbb{R}$ by $f(x) = y$ iff $y \in V$ and $x - y$ is rational. Now for any meager $M \subseteq [0, 1]$, $f[M] \subseteq M + \mathbb{Q}$ is meager. So $f$ has property $\mathcal{M}$. As range of $f$ is $V$, $f$ doesn't send sets with BP to sets with BP. So it doesn't have property $\mathcal{B}$.

Analogous statements hold if you replace "meager" by "Lebesgue null" and "BP" with "Lebesgue measurable".


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