Unit probability subset of image of a measurable set

Question

Let $(\Omega,\Sigma,P)$ be a probability space and $A\in \Sigma$ be such that $P(A) = 1$. Let $X:\Omega \to \mathbb{R}^n$ be a $\Sigma$-measurable map and $\Lambda_X (B) = P(X \in B)~\forall B \in \mathcal{B}(\mathbb{R^n})$. Now, $X(A)$ need not be measurable but does there exist a measurable set $B \subset X(A)$ such that $\Lambda_X(B) = 1$?

This question arose while defining regular conditional distributions given values (i.e, $P(Y \in B | X=x$)) based on regular conditional distributions, given $\sigma$-algebras (i.e, $P(Y \in B | \sigma(X))$. Leo Breiman in his text Probability seems to state that this is straight-forward (paragraph after Proposition 4.35)

Answer 1

It is not true in general. Let $\Omega = V \subset [0,1]$ be a set of outer Lebesgue measure 1 and inner measure $c < 1$ (you may take the complement of a familiar Vitali set of inner measure 0), and let $\Sigma = \{B \cap V : B \in \mathcal{B}([0,1])\}$. Define the measure $P$ on $\Sigma$ by $P(B \cap V) = m(B)$ where $m$ is Lebesgue measure on $[0,1]$. Since $V$ has outer measure 1, this is well defined: if $B,C \in \mathcal{B}(\mathbb{R}^n)$ with $B \cap V = C\cap V$, then $B \triangle C \subset V^c$, so that $m(B \triangle C) = 0$ and therefore $m(B) = m(C)$.

Now let $X : V \to [0,1]$ be the inclusion map, and take $A=V$. If $B \in \mathcal{B}([0,1])$ with $B \subset X(A) =V$, then $m(B) \le c < 1$ (since $V$ has inner measure $c$) and $$\Lambda(B) = P(X^{-1}(B)) = P(B \cap V) = m(B) \le c <1.$$

But it is true if $(\Omega, \Sigma)$ is standard Borel, i.e. $\Omega$ is a Polish space and $\Sigma$ is its Borel $\sigma$-algebra.

Let $\Omega' = \Omega \times \mathbb{R}^n$ with its product topology, and $\Sigma'$ the Borel $\sigma$-algebra of $\Omega_1$. Define $X' : \Omega \to \Omega'$ by $X'(\omega) = (\omega, X(\omega))$ and let $\Lambda' = P \circ X'^{-1}$ be the pushforward of $P$ under $X'$. Also let $\pi : \Omega' \to \mathbb{R}^n$ be the projection onto the $\mathbb{R}^n$ coordinate. Note that $\pi$ is continuous and $X = \pi \circ X'$.

Since $X'$ is Borel and injective, you can show that $X'(A)$ is Borel in $\Omega_1$. (The graph of $X$ is Borel in $\Omega'$ - easy exercise or see Srivastava, A Course in Borel Sets, Proposition 3.1.21 - and $X'(A)$ is the intersection of the graph with the Borel set $A \times \mathbb{R}^n$.) Clearly $\Lambda'(X'(A)) = 1$. Now the measure $P'$, being a Borel probability measure on a Polish space, is Radon, so $\Lambda'(A)$ can be approximated from within by compact sets $K_n' \subset X'(A)$ such that $\Lambda'(K_n') \to 1$ [Srivastava Theorem 3.4.19]. Then $K_n = \pi(K_n')$ is compact in $\mathbb{R}^n$, hence measurable, and we have $K_n \subset X(A)$ with $\Lambda(K_n) = \Lambda'(K_n') \to 1$. So letting $B = \bigcup_n K_n$ we have the desired set.

As a more sophisticated fact, if $f : Y \to Z$ is a Borel map between standard Borel spaces, then the image of any Borel set in $Y$ is analytic in $Z$, and any analytic set in a standard Borel space is universally measurable.