Is the product $d x \otimes \mu_x(d y)$ on $[0,1] \times [0,1]$ a probability measure? Let $Y=[0,1]$ and $X=[0,1]$ (both with the usual Borel sigma algebra).
Let $\mu_x$ be a probability measure over $Y=[0,1]$ for each $x \in X$. (i.e $\mu_x$ depends on $x \in X$).
Let $d x$ be the usual Lebesgue measure on $X=[0,1]$.
Moreover let $X \ni x \rightarrow \int_{[0,1]} f(y) \mu_x (d y) \in \mathbb R$ be continuous for each $f \in C([0,1],\mathbb R)$?
Question:
Is $\mu(d x, dy) = d x \otimes \mu_x(d y)$ a probability measure on $[0,1] \times [0,1]$?
If not: What assumption on $\mu_x$ would be sufficient?
 A: Since $x \mapsto \int_{[0,1]} f(y)\mu_x(dy)$ is continuous in $x$ for each continuous $f$, we can easily extend to say that $x \mapsto \int_{[0,1]} f(y)\mu_x(dy)$ is borel for all positive borel $f$ (take increasing limits).
Thus
$$
I_f(x) := \int_{[0,1]} f(y)\mu_x(dy)
$$
is borel for each borel positive $f$. We can then define the kernel
$$
k(x,A) := I_{1_A}(x) = \int_{A} \mu_x(dy).
$$
We call it a kernel because for every $x$, $A\mapsto k(x,A)$ is a probability measure, and for every $A$ borel, $x\mapsto k(x,A)$ is measurable.
Whenever we have a kernel, we can define a measure $\mu$ via
$$
\int_{A\times B} d\mu := \mu(A \times B) = \int_B k(x,A)\nu(dx)
$$
for all $A,B$ borel for another borel measure $\nu$ (in your case Lebesgue). (Actually we can use any measure space $(E,\mathscr{E},\nu)$, but let's just stick to borel). We have just defined $\mu$ on measurable rectangles $A\times B$, but we can extend by linearity and monotone convergence to the entire product space. The resulting measure will be unique if it is $\sigma$-finite. It's total measure is
$$
\mu([0,1]\times[0,1]) = \int_{[0,1]} k(x,[0,1])dx = \int_{[0,1]} 1 dx = 1
$$
so indeed it is $\sigma$-finite, and hence unique, and moreover a probability measure. No extra conditions on $\mu_x$ were needed.
