Let $(S,\mathcal{S})$ and $(T,\mathcal{T})$ be measurable spaces and consider a measurable function $\phi: S\to T$. Define a probability kernel $\Phi$ from $S$ to $T$ by $\Phi(x,\cdot) = \delta_{\phi(x)}$.
Suppose there is a probability kernel $\lambda$ from $T$ to $S$ such that $$\lambda \Phi(y,\cdot) = \delta_y.$$
Show that for bounded measurable functions $f:T\to \mathbb{R}$ and $g:S\to \mathbb{R}$, $$\lambda(\Phi f)g = f\lambda g.$$
I don't see why $\lambda(\Phi f)g = f\lambda g$. What I did was the following:
$$\lambda(\Phi f)(x) = (\lambda \Phi)(f)(x) = \int_S \lambda \Phi(x,dz) f(z) = \int_S \delta_{\phi(x)}(dz) f(z)$$
And now I'm stuck. Does anyone know what I'm doing wrong here?
Question. Let $\big( S, \mathcal{S} \big)$ and $\big( T, \mathcal{T} \big)$ be two measurable space and $\phi : S\to T$ be a measurable function. Define a kernel $\Phi$ from $S$ to $T$ by $\Phi(s, A \big) = 1_{\phi(s)\in A}$ for $s\in S$ and $A\in\mathcal{T}$. In other words, $\Phi$ maps $S\times\mathcal{T}$ to $[ 0,1]$ such that for any $s\in S$, $\Phi(s, \cdot )$ is a probability measure on $\big( T, \mathcal{T} \big)$ and for any $A\in\mathcal{T}$, $s\in S\longmapsto \Phi(s, A)\in [0,1]$ is $\mathcal{S}/\mathscr{B}(\mathbb{R})$-measurable.
Let $\Lambda$ be a Kernel from $T$ to $S$ such that $\Lambda\Phi$ is an identity Kernel on $T$, that is, for $t\in T$ and $A\in\mathcal{T}$, (see page 80, Chapter III, Revuz et Yor.) $$\Lambda\Phi( t, A ) : = \int_S \Phi(s, A)\, \Lambda(t, ds) = \delta_t(A)\,\, . \quad(\star) $$ Show that for any bounded measurable functions $f : T\to \mathbb{R}$ and $g : S\to \mathbb{R}$, we have $\Lambda\big[ \big(\Phi f\big)\cdot g \big] = f \big[ \Lambda g \big]$.
Proof. Clearly for $s\in S$, (the following is defintion of $\Phi f$.) $$\Phi f(s) = \int_T f(t) \, \Phi(s, dt) = \int_T f(t) \,\delta_s(dt) = f\big(\phi(s) \big)\in\mathbb{R} $$
Then $(\Phi f)g$ is a product of two bounded measurable functions on $S$, hence is a bounded measurable function on $S$.
In this question, $\Phi(s, A) = \delta_{\phi(s)}(A)$, by $(\star)$, $$\Lambda\Phi( t, A ) : = \int_S \textbf{ 1}_{A}\big(\phi(s) \big)\, \Lambda(t, ds) = \int_{\phi^{-1}(A)} \, \Lambda(t, ds) = \Lambda\big(t , \phi^{-1}(A) \big) = \delta_t(A) $$
$\Big($ $\Lambda(t, ds)$ is a probability measure on $\big( S, \mathcal{S} \big)$, $\Lambda\big(t , \phi^{-1}(A) \big)$ can be viewed as the pull back of this probability measure by $\phi^{-1}$, just like $\mathbb{P}\circ\phi^{-1}(A) = \mathbb{P}\Big[ \phi^{-1}(A) \Big]$ . $\Big)$
Then the desired result follows: by a change of variable $u = \phi(s)$ \begin{align*} \Big[ \Lambda(\Phi f)g \Big](t) & = \int_S f\circ\phi(s)\cdot g(s)\, \Lambda(t, ds) = \int f(u) \cdot g(\phi^{-1}(u))\, \delta_{t}(du) \\ & = f(t) \int g(\phi^{-1}(u))\, \delta_{t}(du) \\ & = f(t) \int_S g(s)\,\Lambda(t, ds) = f(t)\cdot \big[ \Lambda g \big](t) \end{align*}
Q.E.D.