Restriction of adjoint map on measure space I am stuck at a certain computation in solving a problem. The problem can be formulated as the following:
Let $X, Y$ be locally compact spaces and $C_b(X)$ denotes the space of bounded continuous functions on $X$ and $M(X)$ denotes the space of regular complex Borel measures on $X$. 
We have a linear map  $T: C_b(Y) \rightarrow C_b(X)$. Then we can get the adjoint map $T^*: C_b(X)^* \rightarrow C_b(Y)^*$ given by $$T^*(m)(f) = m(Tf)$$
for each $m\in C_b(X)^*, f\in C_b(Y)$.
We know that $M(X) \subset C_b(X)^*$. My question is the following:
If for every $x\in X$, we have that $$T^*(\delta_x) = \delta_{x'}$$ for some $x'\in Y$ (here $\delta_x$ is the Dirac measure at the point $x$), then can we conclude that
$$T^*({M(X)}) \subset M(Y) \ \ ?$$
I believe it should be true since $C_0(X)^* = M(X)$ and since $C_0(X) \subset C_b(X) = C(\tilde{X})$ where $\tilde{X}$ is the one-point compactification of $X$. 
But I am not sure if this is the correct line of thought or if I am missing something here. Any help is greatly appreciated !
 A: Yes, I think it's true.
If for all $x$ there exists (unique) $y$ such that $T^* \delta_x = \delta_{y}$, then let $\phi : X \to Y$ be the map $x \mapsto y$.  Then it is clear that $T = \phi^*$ is the pullback operator induced by $\phi$: $Tf = f \circ \phi$.  Since $X,Y$ are locally compact Hausdorff (I am guessing you mean your spaces to be Hausdorff) and hence completely regular, it follows that $\phi$ is continuous.  
(To see this, let $E \subseteq Y$ be closed.  By complete regularity, for each $y \in Y \setminus E$ there is a bounded continuous $f_y : Y \to \mathbb{R}$ with $f_y(E) = 0$ and $f_y(y)=1$.  Hence $E = \bigcap_y f_y^{-1}(0)$.  Now $\phi^{-1}(E) = \bigcap_y \phi^{-1}(f_y^{-1}(0)) = \bigcap_y (f_y \circ \phi)^{-1}(0)$.  By assumption each $f_y \circ \phi = T f_y$ is continuous so $\phi^{-1}(E)$ is closed.)
Now $T^*= \phi_*$ is the pushforward operator induced by $\phi$.  It suffices (by Jordan decomposition) to show that $T^*$ takes positive finite regular measures to positive finite regular measures, so let $\mu$ be a positive finite regular measure on $X$.  Since $\phi$ is continuous and hence Borel measurable, it is clear that $\nu = T^* \mu = \phi_* \mu = \mu \circ \phi^{-1}$ is a positive finite measure on $X$.  
To see that $\nu$ is inner regular, let $B \subseteq Y$ be Borel and fix $\epsilon > 0$.   Now $A = \phi^{-1}(B)$ is Borel in $X$, and by definition $\nu(B) = \mu(A)$.  By regularity of $\mu$, there is a compact $K \subseteq A$ with $\mu(K) \ge \mu(A) - \epsilon$.  Then $F = \phi(K)$ is compact in $X$, and $F \subseteq \phi(A) = B$.  On the other hand, $\phi^{-1}(F)$ is closed in $X$ and is a superset of $K$, so $$\nu(F) = \mu(\phi^{-1}(F)) \ge \mu(K) \ge \mu(A) - \epsilon = \nu(B) - \epsilon.$$
Outer regularity of $\nu$ now follows by taking complements, since $\nu$ is a finite measure.
