For a Borel function when does there exist a set of full measure with measurable image

Question

Let $(X,\Sigma,\mu)$ be a standard Borel probability space. Consider a measurable and measure preserving self map $f:X\to X$. We know that $f(A)$ is analytic for any $A\in \Sigma$, but this does not necessarily mean that $f(A)$ is Borel measurable.

What are the weakest conditions we can impose on $X$ or $f$ to ensure the existence of some $A\in \Sigma$, such that $f(A)\in \Sigma$ and $\mu(A)=1$?

I say weakest, because we know that if $f$ is injective then $X$ would suffice (because if the complement of an analytic set is analytic then it is Borel).

EDIT: Michael's comment gives a very useful approach. I think surjectivity is still quite a strong condition, but if there was a weaker condition we could impose ensuring that the image contains a measurable set of full measure that would be grand.

Answer 1

Here is the general case (for possibly incomplete measure spaces). Recall that Property P is defined in my other answer.

Let $(X, \Sigma, \mu)$ be the original measure space. Let $(X, \Sigma_c, \mu_c)$ be the completed measure space. Note that $\mu(B)=\mu_c(B)$ for all $B \in \Sigma$.

Claim:

Property P holds if and only if $f(X) \in \Sigma_c$.

To prove the claim we first have some lemmas.

Lemma 1:

Every element of $\Sigma_c$ can be written as $R \cup H$ where $R \in \Sigma$ and $H$ is a subset of a measure-0 set in $\Sigma$.

Proof: (see wikipedia for this lemma here): https://en.wikipedia.org/wiki/Complete_measure

Lemma 2:

If $f$ is measurable and measure-preserving under $(X, \Sigma, \mu)$, then it is also measurable and measure-preserving under $(X, \Sigma_c, \mu_c)$.

Proof: Let $S \in \Sigma_c$. We want to show $f^{-1}(S) \in \Sigma_c$ and $\mu_c(f^{-1}(S)) = \mu_c(S)$.

We know $S= R \cup H$ for $R \in \Sigma$ and $H \subseteq Z$ for some $Z \in \Sigma$ with $\mu(Z)=0$. So $$f^{-1}(S) = f^{-1}(R \cup H) = f^{-1}(R) \cup f^{-1}(H) \quad (*) $$
But $f^{-1}(H) \subseteq f^{-1}(Z)$ and $\mu(f^{-1}(Z))=\mu(Z)=0$ by the measure-preserving property. So $f^{-1}(H)$ is a subset of a measure zero set and so $f^{-1}(H) \in \Sigma_c$. Also, $f^{-1}(R) \in \Sigma \subseteq \Sigma_c$. Thus, (*) shows that $f^{-1}(S)$ is the union of two sets in $\Sigma_c$, so $f^{-1}(S) \in \Sigma_c$.

Next, note that since $S = R \cup H$ with $\mu_c(H)=0$, we know $$\mu_c(S)=\mu_c(R) \quad (**) $$

On the other hand, since (* ) writes $f^{-1}(S)$ as the union of two sets in $\Sigma_c$ we get:
\begin{align} \mu_c(f^{-1}(S)) &\overset{(a)}{\leq} \mu_c(f^{-1}(R)) + \mu_c(f^{-1}(H)) \\ &\overset{(b)}{\leq} \mu_c(f^{-1}(R)) + \mu_c(f^{-1}(Z)) \\ &\overset{(c)}{=} \mu(f^{-1}(R)) + \mu(f^{-1}(Z))\\ &\overset{(d)}{=} \mu(R) + \mu(Z) \\ &\overset{(e)}{=} \mu(R) + 0 \\ &\overset{(f)}{=} \mu_c(R) \\ &\overset{(g)}{=} \mu_c(S) \end{align} where (a) holds by (*); (b) holds because $H\subseteq Z$; (c) holds because $R$ and $Z$ are both in $\Sigma$; (d) holds because $f$ is measure-preserving in the original measure space; (e) holds because $\mu(Z)=0$; (f) holds because $R \in \Sigma$; (g) holds by (**). $\Box$

Proof of claim that Property P holds if and only if $f(X) \in \Sigma_c$:

Proof ($\Longrightarrow$): Suppose Property P holds. Then there is a set $A$ such that $A \in \Sigma$, $\mu(A)=1$, and $f(A) \in \Sigma$. Since every element of $\Sigma$ is also in $\Sigma_c$, and $1=\mu(A)=\mu_c(A)$, it follows that property P holds in the completed measure space. Since $f$ is measurable and measure-preserving in the completed measure space, and since property P holds in the completed measure space, by Claim 2 in my previous answer it follows that $f(X) \in \Sigma_c$. $\Box$

Proof ($\Longleftarrow$): Suppose $f(X) \in \Sigma_c$. Then $f(X) = R \cup H$ where $R \in \Sigma$ and $H\subseteq Z$ where $Z \in \Sigma$ and $\mu(Z)=0$. Then $\mu_c(H)=0$ and so $\mu_c(f(X)) = \mu_c(R) = \mu(R)$.

However, we have $f^{-1}(f(X)) = X$, and since $f$ is measure-preserving in the completed measure space we have
$$1=\mu(X)=\mu_c(X) = \mu_c(f^{-1}(f(X)) = \mu_c(f(X)) = \mu(R)$$
So define $A = f^{-1}(R)$. Since $R \in \Sigma$ and $f$ is measurable, we know $A \in \Sigma$. Since $f$ is measure-preserving we know $\mu(A)=\mu(f^{-1}(R))=\mu(R)=1$. Finally, we know $f(A)=R \in \Sigma$. Thus, Property P holds in the original measure space. (To show that $f(A)=R$ note that $f(A) = f(f^{-1}(R)) \subseteq R$. However, since $R\subseteq f(X)$, every element of $R$ can be mapped to by $f$, so $f(f^{-1}(R))=R$.) $\Box$

Answer 2

My comments above assumed a general measure space with $\mu(X)$ not necessarily 1. The case $\mu(X)=1$ is a bit easier.

Assume:

• $X$ is a measure space with sigma algebra $\Sigma$ and measure $\mu$.

• $\mu(X)=1$

• $f:X\rightarrow X$ is a measurable and measure-preserving function.

Definition:

Let's say that Property P holds if there is a set $A \in \Sigma$ such that $\mu(A)=1$ and $f(A) \in \Sigma$.

Claim 1:

Suppose $f(X) \in \Sigma$. Then Property P holds.

Proof: Define $A=X$. Then $A \in \Sigma$, $\mu(A)=1$, and $f(A) = f(X) \in \Sigma$. So Property P holds. $\Box$

Claim 2:

Suppose the sigma algebra is complete, so that all subsets of measure-0 sets also have measure 0. Then Property P holds if and only if $f(X) \in \Sigma$.

Proof ($\Longleftarrow$): Suppose $f(X) \in \Sigma$. Then Claim 1 ensures Property P holds. $\Box$.

Proof ($\Longrightarrow$): Suppose Property P holds. Let $A$ be a set such that $A \in \Sigma$, $\mu(A)=1$, $f(A)\in \Sigma$. We know: $$ A \subseteq f^{-1}(f(A)) \quad (*) $$ and since $f(A)$ is measurable we have $f^{-1}(f(A))$ is measurable and $$ 1= \mu(A) \overset{(a)}{\leq} \mu(f^{-1}(f(A))) \overset{(b)}= \mu(f(A)) \overset{(c)}{\leq} 1$$ where (a) holds by (*); (b) holds by the measure-preserving property of $f$; (c) holds because $f(A) \subseteq X$ and $\mu(X)=1$. It follows that $\mu(f(A))=1$ and so $$ \mu(f(A)^c) = 0$$ But $f(X)^c \subseteq f(A)^c$ and so $\mu(f(X)^c)=0$ (by completeness). Thus, $f(X)^c \in \Sigma$ and so $f(X)\in \Sigma$. $\Box$

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Examples

1) If we use $X=\{1, 2, 3, 4\}$, $\Sigma = \{\phi, X, \{1, 2\}, \{3, 4\} \}$, $f(1)=f(2)=2, f(3)=f(4)=4$, $\mu(\{1,2\})=\mu(\{3,4\})=1/2$, then $\mu(X)=1$, $f$ is measurable and measure-preserving, this is a complete measure space, but Property P does not hold because $f(X) = \{2, 4\} \notin \Sigma$.

2) Same $X$ and $\Sigma$ as before, but define $f:X\rightarrow X$ by $f(1)=1, f(2)=2, f(3)=4, f(4)=4$. Define $\mu(\{1,2\})=1, \mu(\{3,4\})=0$. This is a non-complete measure space with $\mu(X)=1$, $f$ is measurable and measure-preserving, and Property P holds with respect to the set $A = \{1, 2\}$. However, $f(X)$ is not measurable.

Now, if we complete the measure by adding the null sets $\{3\}$ and $\{4\}$ to $\Sigma$, then indeed $f(X)$ is measurable with measure 1.