Let $\nu$ be a probability measure equivalent to $\mathbf{1}_{\mathbb{R}_+}(y) \, \lambda(dy)$. Let $\pi$ be a probability measure on $\mathbb{R}^2$ of second marginal $\nu$, such that $\nu(dy)$-a.e., $$ \pi_{Y = y}(dx) = \frac{\mathbf{1}_{A_y}(x)}{\lambda(A_y)} \, dx, $$ with $A_y \;=\; \bigl[\{y\} - 2^{-\lfloor y \rfloor},\;\{y\} + 2^{-\lfloor y \rfloor}\bigr] \cap [0,1]$, where $\{y\}$ denotes the fractional part of $y$ and $\lfloor y \rfloor$ denotes the floor of $y$. Since $\mu$ is absolutely continuous with respect to Lebesgue measure $\lambda$ on $[0,1]$, there exists a density $m : [0,1] \to \mathbb{R}_+$ such that $$ \mu(dx) \;=\; m(x)\,\lambda(dx), $$ the conditional law of $Y$ given $X=x$ admits the following expression: $$ \pi_{X=x}(dy) \;=\; \frac{\mathbf{1}_{A_y}(x)/\lambda(A_y)}{m(x)}\,\nu(dy), \qquad \text{for $\mu$-a.e. } x\in[0,1]. $$
The operator $$ T : L^1(\nu) \to L^1(\mu), $$ \begin{gather*} Tg(x) = \int_{y \in \mathcal{Y}} g(y) \, \pi_{X=x}(dy) \\ = \frac{1}{m(x)} \int_{\mathcal{Y}} g(y)\,\frac{\mathbf{1}_{\bigcup_{n \in \mathbb{N}} \bigl([n + x - 2^{-n},\, n + x + 2^{-n}] \cap [n, n+1)\bigr)}(y)}{\lambda(A_y)}\,\nu(dy) =\mathbb{E}[g(Y) | X = x]. \label{eq:def_T_operator} \end{gather*} is surjective.
I show the surjectivity by showing that the operator adjoint $T^*$ defined as $$T^* : L^\infty(\mu) \to L^\infty(\nu)$$ $$T^*f(y) = \int_{x \in \mathbb{R}} f(x) \, \pi_{Y=y} (dx)= \mathbb{E}[f(X)|Y=y]$$ is lower bounded (proof below).
My question: I wonder if we can be more explicit and for every function $f$ in $L_1(\mu)$, construct a function $g$ in $L_1(\nu)$, such that $Tg=f$. This amount to solving explicitly
$$ \mathbf{1}_{[0,1]}\lambda(dx)\text{-a.e., }\quad f(x) = \int_{y \in \mathbb{R}^+} g(y)\, \mathbf{1}_{\bigcup_{n \in \mathbb{N}} [n + x - 2^{-n},\, n + x + 2^{-n}] \cap [n, n+1)}(y)\, \lambda(dy). $$
Proof of surjectivity
Let $\delta\in(\frac{1}{2},1)$. We are going to check that $
\nu\bigl(\{\,y \in \mathbb{R}_+:\,\pi_{Y=y}(A)>\delta\}\bigr)>0
$ for all $A\in \mathcal{B}(\mathbb{R})$ such that $\mu(A)>0$. Let $A\in\mathcal{B}(\mathbb{R})$ be such that $\mu(A)>0$. Without loss of generality, we may assume that $A \in \mathcal{B}([0,1])$, since $\operatorname{supp}(\mu) \subset [0,1]$. By the Lebesgue differentiation theorem, we have
$$
\mathbf{1}_A(x)\, \lambda(dx) \text{-a.e., } \quad
\frac{\lambda(A \cap [x-\varepsilon, x])}{\lambda([x-\varepsilon, x])}
\xrightarrow[\varepsilon \to 0^+]{} 1,\quad
\frac{\lambda(A \cap [x, x+\varepsilon))}{\lambda([x, x+\varepsilon))}
\xrightarrow[\varepsilon \to 0^+]{} 1.
$$
Hence, there exist $x\in(0,1)$ and $\varepsilon>0$ such that $[x-\varepsilon,x+\varepsilon]\subset[0,1]$ and
$$
\forall \varepsilon' \in (0, \varepsilon), \quad
\frac{\lambda\bigl(A\cap[x-\varepsilon',x]\bigr)}{\varepsilon'}
>\delta,
\quad
\frac{\lambda\bigl(A\cap[x,x+\varepsilon']\bigr)}{\varepsilon'}
>\delta.
$$
In particular, for every interval $I$ of positive length, such that $I \subset [x-\varepsilon, x+\varepsilon]$ and $x \in I$, we have $$
\frac{\lambda\bigl(A\cap I\bigr)}{\lambda(I)} > \delta.
$$
Let $n\in\mathbb{N}$ and $y\in[n,n+1)$.
We observe that
$$
y \in \left[ n + x - \min\left(2^{-n},\, \varepsilon - 2^{-n} \right),\
n + x + \min\left(2^{-n},\, \varepsilon - 2^{-n} \right) \right]
\quad \text{and} \quad 2^{-n} \leq \varepsilon
$$
implies that
$$
A_y \subset [x - \varepsilon,\, x + \varepsilon]
\quad \text{and} \quad x \in A_y.
$$
Thus for $n\geq \lceil-\log_2\varepsilon\rceil +1$, we have $2^{-n}\le \frac{e^{-\lceil-\log_2\varepsilon\rceil\log 2}}2\le \frac{\varepsilon}2$, so that
$$
[n+x-2^{-n},n+x+2^{-n}] \subset \{\,y\in[n,n+1):A_y\subset[x-\varepsilon,x+\varepsilon] \text{ and } x\in A_y \}.
$$
Therefore, by hypothesis on $\nu$, we have
$
\nu\left(\left\{\, y \in [n, n+1) : A_y \subset [x - \varepsilon, x + \varepsilon]\text{ and } x\in A_y \,\right\}\right)> 0.
$
It follows that
$$
\nu\left(\left\{\,y \in \mathbb{R}_+:\,\pi_{Y=y}(A) > \delta\,\right\}\right)
=
\nu\left(\left\{\,y \in \mathbb{R}_+:\,\frac{\lambda(A \cap A_y)}{\lambda(A_y)} > \delta\,\right\}\right)
\ge
\nu\left(\left\{\,y \in \mathbb{R}_+:\, A_y \subset [x - \varepsilon, x + \varepsilon] \text{ and } x\in A_y \right\}\right)
> 0.$$
Let $ f^* $ be a simple function in $ L^\infty(\mu) $.
There exists a finite index set $ I $, a collection of disjoint Borel sets $ (A_i)_{i \in I} $ with $ \mu(A_i) > 0 $,
and a vector $ (f^*_i)_{i \in I}\in \mathbb{R}^I$ such that
$$
\forall x \in \mathbb{R},\text{ }f^*(x) = \sum_{i \in I} f^*_i\, \mathbf{1}_{A_i}(x).
$$
Let $ i_0 \in I $ such that $\max_{i \in I} \lvert f_i^* \rvert = \lvert f^*_{i_0} \rvert = \lVert f^*\rVert_{L^\infty(\mu)}$.
We have $\nu(dy)$-a.e.,
\begin{align*}
\left| T^*f^*(y) \right| &= \left| \sum_{i \in I} f^*_i \, \pi_{Y=y}(A_i) \right| \\
&\geq \left| f^*_{i_0} \right| \pi_{Y=y}(A_{i_0}) - \sum_{i \in I \setminus \{i_0\}} \left| f^*_{i_0} \right| \pi_{Y=y}(A_i) \\
&= \left( 2\, \pi_{Y=y}(A_{i_0}) - 1 \right) \left| f^*_{i_0} \right|.
\end{align*}
Thus $T^*$ is lower-bounded, so that $T$ is surjective.
