To what extent does conditional distribution determine conditional expectation? Suppose $P\left(\left.X\right|\mathcal{A}\right)$ and $P\left(\left.Y\right|\mathcal{A}\right)$ are regular conditional distributions that satisfy: for any Borel set $B$, $P\left(\left.X\in B\right|\mathcal{A}\right)=P\left(\left.Y\in B\right|\mathcal{A}\right)\space\mathrm{a.s.}$ Does this imply that $E\left(\left.f(X)\right|\mathcal{A}\right)=E\left(\left.f(Y)\right|\mathcal{A}\right)\space\mathrm{a.s.}$ for all integrable $f$?
Since $E\left(\left.f(X)\right|\mathcal{A}\right)\left(\omega\right)=\int f dP\left(\left.\cdot\right|\omega\right)\space\mathrm{a.s.}$ and likewise for $E\left(\left.f(Y)\right|\mathcal{A}\right)$, the answer would be affirmative if it can be shown that for almost all $\omega$, $P\left(\left.X\in\cdot\right|\mathcal{A}\right)\left(\omega\right)=P\left(\left.Y\in\cdot\right|\mathcal{A}\right)\left(\omega\right)$, but is this the case?
 A: Following Ilya's tip in a comment to the original post, here's a fuller answer.
Let $S:=\left(\Omega,\mathcal{F},P\right)$ be a probability space, let $\mathcal{A}$ be a sub-$\sigma$-algebra of $\mathcal{F}$ and set $S':=\left(\Omega,\mathcal{A},P\right)$. Also let $X_1,X_2:\Omega\rightarrow\mathbb{R}$ be $\mathcal{F}\space/\space\mathfrak{B}$ measurable ($\mathfrak{B}$ being the standard Borel field on the real line).
Now let $\kappa_1,\kappa_2:\mathfrak{B}\times\Omega\rightarrow\left[0,1\right]$ be regular versions of the conditional distributions $P\left(\left.X_1\in\cdot\right|\mathcal{A}\right)$, $P\left(\left.X_2\in\cdot\right|\mathcal{A}\right)$, respectively. In other words, assume $\kappa_i$ ($i\in\left\{1,2\right\}$) satisfy the following two conditions:


*

*For all $\omega\in\Omega$, the function
$$\mu_\omega^{(i)}:\mathfrak{B}\rightarrow\mathbb{R},\space\space\mu_\omega^{(i)}(D):=\kappa_i\left(D,\omega\right)$$
is a probability measure on $\left(\mathbb{R},\mathfrak{B}\right)$.

*For all $D\in\mathfrak{B}$, the function
$$\phi_D^{(i)}:\Omega\rightarrow\left[0,1\right]\space\space\phi_D^{(i)}\left(\omega\right):=\kappa_i\left(D,\omega\right)$$
is a version of the conditional probability $P\left(\left.X_i\in D\right|\mathcal{A}\right)$

*Suppose, in addition, that for all $D\in\mathfrak{B}$, $\phi_D^{(1)}=\phi_D^{(2)}\space\space S'\mathrm{-a.s.}$.
It is required to show that under these circumstances, for all $\mathfrak{B}\space/\space\mathfrak{B}$-measurable $f:\mathbb{R}\rightarrow\mathbb{R}$ s.t. $f\circ X_1, f\circ X_2\in\mathcal{L}_1(S)$, if $m_i:\Omega\rightarrow\mathbb{R}$ ($i\in\left\{1,2\right\}$) is a version of $E_{S'}\left(\left.f\circ X_i\right|\mathcal{A}\right)$, then $m_1=m_2\space\space S'\mathrm{-a.s.}$.
Let then $f:\mathbb{R}\rightarrow\mathbb{R}$ s.t. $f\circ X_1, f\circ X_2\in\mathcal{L}_1(S)$ and let  $m_i:\Omega\rightarrow\mathbb{R}$ ($i\in\left\{1,2\right\}$) each be a version of $E_{S'}\left(\left.f\circ X_i\right|\mathcal{A}\right)$.
$f$ falls into one of the following four (not mutually exclusive) categories:
A. $f$ is an indicator, i.e. $f=\mathbb{1}_D$ for some $D\in\mathfrak{B}$.
B. $f$ is a non-negative linear combination of indicators: $f=\sum_{j=1}^n c_j\mathbb{1}_{D_j}$ for some $1\leq n\in\mathbb{N}$, $0\leq c_j\in\mathbb{R}$ and $D_j\in\mathfrak{B}$, $j\in\left\{1,\dots,n\right\}$.
C. $f$ is non-negative.
D. $\left(f\circ X_i\right)^\pm\in\mathcal{L}_1(S)$ ($i\in\left\{1,2\right\}$).
We will proceed inductively by considering each category in turn and showing that if it is true that $m_1=m_2$ $S'$-a.s. given that $f$ belongs to the previous category, then it is true too if $f$ belongs to the current category.
A) Base case: assume that $f=\mathbb{1}_D$ for some $D\in\mathfrak{B}$. Then $m_i$ is a version of $E_{S'}\left(\left.\mathbb{1}_D\circ X_i\right|\mathcal{A}\right)=P\left(\left.X_i\in D\right|\mathcal{A}\right)$. Then by condition 2, $m_i=\phi_D^{(i)}$ $S'$-a.s. But by condition 3, $\phi_D^{(1)}=\phi_D^{(2)}$ $S'$-a.s.
B) Assume that $f=\sum_{j=1}^n c_j\mathbb{1}_{D_j}$ for some $1\leq n\in\mathbb{N}$, $0\leq c_j\in\mathbb{R}$ and $D_j\in\mathfrak{B}$, $j\in\left\{1,\dots,n\right\}$. So $f\circ X_i=\sum_{j=1}^n c_j\left(\mathbb{1}_{D_j}\circ X_i\right)$ ($i\in\left\{1,2\right\}$) and so, based on the previous case and on the linearity of conditional expectation,
$$E_{S'}\left(\left.f\circ X_1\right|\mathcal{A}\right)=\sum_{j=1}^n c_j E_{S'}\left(\left.\mathbb{1}_{D_j}\circ X_1\right|\mathcal{A}\right)=\sum_{j=1}^n c_j E_{S'}\left(\left.\mathbb{1}_{D_j}\circ X_2\right|\mathcal{A}\right)=E_{S'}\left(\left.f\circ X_2\right|\mathcal{A}\right)\space\space S'\mathrm{-a.s.}$$
C) Assume that $f$ is non-negative. Then $f$ is the monotone limit $f_k\uparrow f$ of some sequence of functions $\left(f_k\right)_{k=1}^\infty$ where for each $k\in\mathbb{N}$, $f_k$ is a non-negative linear combination of indicators. Hence $f_k\circ X_i\uparrow f\circ X_i$ ($i\in\left\{1,2\right\}$) and so, based on the previous case and on the monotone convergence property of conditional expectation,
$$\begin{array} {lcl}
E_{S'}\left(\left.f\circ X_1\right|\mathcal{A}\right) & = & E_{S'}\left(\left.\lim_{k\rightarrow\infty}\left(f_k\circ X_1\right)\right|\mathcal{A}\right)=\lim_{k\rightarrow\infty}E_{S'}\left(\left.f_k\circ X_1\right|\mathcal{A}\right) = \\
& = & \lim_{k\rightarrow\infty}E_{S'}\left(\left.f_k\circ X_2\right|\mathcal{A}\right)=E_{S'}\left(\left.\lim_{k\rightarrow\infty}\left(f_k\circ X_2\right)\right|\mathcal{A}\right) = E_{S'}\left(\left.f\circ X_2\right|\mathcal{A}\right)\space\space S'\mathrm{-a.s.}
\end{array}$$
D) Noting that $\left(f\circ X_i\right)^\pm=f^{\pm}\circ X_i$ ($i\in\left\{1,2\right\}$), we get, based on the previous case and on the linearity of conditional expectation,
$$\begin{array}
&& E_{S'}\left(\left.f\circ X_1\right|\mathcal{A}\right) \\
& = & E_{S'}\left(\left.\left(f\circ X_1\right)^{+}-\left(f\circ X_1\right)^{-}\right|\mathcal{A}\right)=E_{S'}\left(\left.\left(f\circ X_1\right)^{+}\right|\mathcal{A}\right)-E_{S'}\left(\left.\left(f\circ X_1\right)^{-}\right|\mathcal{A}\right) \\
& = & E_{S'}\left(\left.f^{+}\circ X_1\right|\mathcal{A}\right)-E_{S'}\left(\left.f^{-}\circ X_1\right|\mathcal{A}\right)=E_{S'}\left(\left.f^{+}\circ X_2\right|\mathcal{A}\right)-E_{S'}\left(\left.f^{-}\circ X_2\right|\mathcal{A}\right) \\
& = & E_{S'}\left(\left.\left(f\circ X_2\right)^{+}\right|\mathcal{A}\right)-E_{S'}\left(\left.\left(f\circ X_2\right)^{-}\right|\mathcal{A}\right) = E_{S'}\left(\left.\left(f\circ X_2\right)^{+}-\left(f\circ X_2\right)^{-}\right|\mathcal{A}\right) \\
& = & E_{S'}\left(\left.f\circ X_2\right|\mathcal{A}\right)\space\space S'\mathrm{-a.s.}
\end{array}$$
Q.E.D.

Comments
The proof does not use the regularity of either $\kappa_1$ or $\kappa_2$, i.e. it does not use property $1$ above that $\kappa_i$ ($i\in\left\{1,2\right\}$) is a probability measure in its first argument.
