Let $\Omega$ be compact, and let $\omega^* \in \Omega$ be arbitrary. Let $\Delta (\Omega)$ denote the set of all probability measures over $\Omega$, and endow $\Delta ( \Omega)$ with the topology of weak converge, i.e. $ \int g d\nu_n \to \int g d\nu$, for every continuous linear functional $g$ on $\Omega$.
Let $\mu \in \Delta ( \Omega)$, and define a function
$$ f : \Delta ( \Omega ) \to \Delta ( \Omega \times \Delta (\Omega)), $$
such that $f ( \mu ) := \mu \otimes [\delta_{\omega^*}]$.
This function is continuous (that's what I found in a book, along with the terrible question "why" in between braces!), but I am not sure about why it is continuous.
I have two guesses in order to prove the result.
- We show that $f$ is Borel measurable, hence continuous. In order to prove that $f$ is Borel measurable, we assume that the domain and the codomain are both endowed with their correspoding Borel $\sigma$-algebra. Hence, we let $B$ be an arbitrary element of the product Borel $\sigma$-algebra of $\Delta ( \Omega \times \Delta (\Omega))$, and we conclude that $f^{-1} (B)$ is of course an element of the Borel $\sigma$-algebra of $\Delta ( \Omega)$ (here, $f^{-1}$ should work intuitively as a projection).
- We prove that for an arbitrary sequence $\mu_n \in \Delta (\Omega)$, if $\mu_n \to \mu$ in $\Delta (\Omega)$, then $f(\mu_n) \to f(\mu)$ in $\Delta ( \Omega \times \Delta (\Omega))$. This should work because the definition of topology of weak convergence ensure us that $\mu_n \to \mu$ in $\Delta (\Omega)$. Hence, $f(\mu_n) \to f(\mu)$ in $\Delta ( \Omega \times \Delta (\Omega))$, because $\mu_n$ converges for the previous reason, and $[\delta_{\omega^*}]$ is constant.
Questions:
Are my line of reasoning corrects in both cases?
Is actually possible to come up with an $\epsilon$-$\delta$ proof?
Any feedback is most welcome.
Thank you for your time.