Dirac measure as an image of pushforward map Let $X, Y$ be two compact spaces and let $Q: X\times Y \to X$ be the canonical projection map. Denote by $Q_*: M(X\times Y)\to M(X)$ the pushforward map, where $M(\cdot)$ is the space of probability Radon measures. Let $Z$ be a subset of $M(X\times Y)$ and assume that $\delta_x\in Q_*(Z)$, where $\delta_x$ denotes the point mass measure. Then, I want to show that there exists a measure $\nu\in M(Y)$ s.t. $\delta_x\otimes \nu\in Z$. I am not sure how to do it since I know that not every measure on a product space is given by a product of measures.
Thanks!
 A: In the following I assume that the spaces are either locally compact Hausdorff (LCH) or Polish, which are pretty much the only settings where these kinds of questions have nice answers.
For convenience I'll denote by $\pi_X : X \times Y \to X$ and $\pi_Y: X \times Y \to Y$ the two projections. 
Let $\mu$ be a measure on $X\times Y$ such that $\pi_{X*} \mu = \delta_x$ for some point $x \in X$. We'll show that in fact $\mu$ is always a product. Intuitively, $\mu$ should somehow be concentrated on the slice $\{x \} \times Y$. This is indeed the case:
Claim: $\mu(U) =0$ for any open set $U$ that doesn't intersect $\{x \} \times Y$.
Proof: Indeed $\pi_X$ is an open map so $\pi_X(U)$ is open (hence measurable) and we have $U \subset \pi_X^{-1}( \pi_X(U))$. Therefore
\begin{equation}
\mu(U) \leq \mu( \pi_X^{-1} ( \pi_X(U))) = \pi_{X*}\mu( \pi_X(U))=\delta_x( \pi_X(U)).
\end{equation}
But $U$ doesn't intersect $\{ x \} \times Y$ so $\pi_X(U)$ doesn't contain $x$, hence $\delta_x( \pi_X(U)) =0$. $\square$
This means that for any open set in $X \times Y$ we have $\mu(U) = \mu ( U \cap \{ x \} \times Y)$. Now suppose that $U$ is a product, that is $U = A \times B$ where $A$ and $B$ are open in $X$ and $Y$ respectively. Then $U \cap \{x \} \times Y = \{x \} \times B$ if $A$ contains $x$, and it is empty otherwise. Hence we have
\begin{equation}
\mu(A \times B) = \delta_x(A) \; \mu ( \{x \} \times B). 
\end{equation}
This means that on products of open sets $\mu$ agrees with the product measure $\delta_x \times \lambda$,
where $\lambda$ is the Borel measure on $Y$ given by 
\begin{equation}
\lambda (B) = \mu ( \{x \} \times B).
\end{equation}
We easily see that $\lambda$ is a probability measure since
\begin{equation}
\lambda(Y) = \mu( \{x \} \times Y) = \mu( \pi_X^{-1}\{ x\} ) = \delta_x( \{x \}) =1.
\end{equation}
In particular, since the product of $\sigma$-finite measures is unique, this implies that
$\mu = \delta_x \times \lambda$ as measures on $B(X) \times B(Y)$ (where $B(\cdot)$ denotes the Borel $\sigma$-algebra). 
If we are working with Polish spaces we are done, since $B(X \times Y) = B(X) \times B(Y)$, and $\lambda$ is automatically Radon (since it is finite). If we are working with LCH spaces we want to prove something more: that $\lambda$ is Radon and that $\mu$ is equal to the product $\delta_x \times \lambda$ on the whole of $B( X \times Y)$, where now $\delta_x \times \lambda$ is interpreted as the Radon product of the measures (this is the unique extension of the usual product on $B(X) \times B(Y)$ to a Radon measure on $B(X \times Y)$).
Let's check that $\lambda$ is a Radon probability measure on $Y$.
$\lambda$ is equal to $f_* \mu|_{\{x \} \times Y}$, where $f: \{x \} \times Y \to Y$ is the obvious homeomorphism. Hence it suffices to show that $\mu|_{\{x \} \times Y}$ is Radon. It is a general fact that the restriction of a $\sigma$-finite Radon measure to a Borel subspace is always Radon. Since $\mu$ is finite, we get that its restriction to $\{x \} \times Y$ is Radon.
Hence $\lambda$ is a Radon probability measure on $Y$. Therefore we can form the Radon product $\delta_x \times \lambda$ on $B(X \times Y)$, as explained earlier. Since $\mu$ is Radon and agrees with $\delta_x \times \lambda$ on products, by the uniqueness of the Radon product we must have $\mu = \delta_x \times \lambda$.  
