Is the expectation operator an open map? Let $\Delta [0,1]$ denote the set of Borel probability measures on $[0,1]$. $\Delta [0,1]$ endowed with the Prokhorov metric is a metric space, as we know. My question is, does the expectation operator map an open set of measures in $\Delta [0,1]$ to an open interval in $[0,1]$? In other words, is expectation an open map? I found this related question, but I'm not able to easily apply related arguments to verify if this claim is true. I'm very new to open maps. Any help is most appreciated.
Edit: In the earlier version of this question I mistakenly said "Borel measures" instead of "Borel probability measures".
 A: Given that you're asking about "expectation" rather than "integral" operators and talking about the Levy-Prokhorov metric, I'm going to assume that you meant that $\Delta[0, 1]$ is the set of probability measures on the Boral $\sigma$-algebra.
In this case, yes, the expectation operator is an open map.
I'll be using the coupling representation of the Prokhorov metric mentioned in this answer: https://math.stackexchange.com/a/4310676/154826
That is,
$$
d_P(\mu, \nu)
= \inf\{\alpha(X, Y) : X \sim \mu, Y\sim \nu\},
$$
where the infimum is taken over all couplings $(X, Y)$, and where
$$
\alpha(X, Y)
= \inf\{\epsilon : \Pr(|X - Y| > \epsilon) \leq \epsilon\}.
$$
We also define the expectation operator $E \colon \Delta[0, 1] \rightarrow [0, 1]$ by $E(\mathcal{L}_X) = \mathbf{E} X$, where $\mathcal{L}_X$ is the law of $X$.
Since all probability measures on $[0, 1]$ can be expressed as laws of random variables, this is well-defined.

So now, in order to demonstrate that $E$ is an open map, we must show that for any open ball $B_\epsilon(\mathcal{L}_X)$, the image $E(B_\epsilon(\mathcal{L}_X))$ contains an open neighbourhood of $E(\mathcal{L}_X) = \mathbf{E} X$.
When $\mathbf{E} X \in (0, 1)$, we will find an open interval
$$
I =\{\kappa \mathbf{E}X : \kappa \in (\kappa_-, \kappa_+)\},
$$
where $\kappa_- < 1 < \kappa_+$ such that $I \subseteq E(B_\epsilon(\mathcal{L}_X))$.
That is, for any $\kappa \mathbf{E} X \in I$, we will show that there is a coupling $(X, Y)$ such that $E(\mathcal{L}_Y) = \mathbf{E} Y = \kappa \mathbf{E} X$ and
$$
\Pr(|X - Y| < \epsilon/2) \leq \epsilon/2
$$
so that $d_P(\mathcal{L}_X, \mathcal{L}_Y) < \epsilon$ and so $\mathcal{L}_Y \in B_\epsilon(\mathcal{L}_X)$.

So let $\kappa \mathbf{E}X \in I$.
If $\kappa <1$, define
$$
Y = \kappa X.
$$
This takes values in $[0, \kappa] \subseteq [0, 1]$, and $\mathbf{E} Y = \kappa \mathbf{E}X$.
Moreover, we have that
\begin{align*}
\Pr(|X - Y| > \epsilon/2)
= \Pr((1 - \kappa) X >\epsilon/2)
= \Pr\biggl(X >\frac{\epsilon/2}{1 - \kappa}\biggr),
\end{align*}
which is $0$ when $\kappa \geq \kappa_- := 1 - 2\epsilon$.
On the other hand, consider the case $\kappa > 1$.
In this case, we define
$$
Y = \lambda + (1 - \lambda)X,
$$
where
$$
\lambda
= \frac{1 - \kappa \mathbf{E} X}{1 - \mathbf{E} X}
= 1 + (1 - \kappa) \frac{\mathbf{E} X}{1 - \mathbf{E} X}.
$$
You can check that this is the linear function that takes $1\mapsto 1$ and $\mathbf{E} X \mapsto \kappa \mathbf{E}X$, and so in particular that $\mathbf{E}Y = \kappa \mathbf{E} X$.
Moreover, we can see that, since $\kappa > 1$, we have that $\lambda \in [0, 1]$, and so $Y$ takes values in $[\lambda, 1] \subseteq [0, 1]$.
Last, we compute
\begin{align*}
\Pr(|X - Y| > \epsilon/2)
= \Pr((1 - \lambda)(1 - X) > \epsilon/2)
= \Pr\biggl(1 - X > \frac{\epsilon/2}{1 - \lambda}\biggr),
\end{align*}
which is $0$ when $1 - \lambda < \epsilon/2$.
But since $\lambda$ is a decreasing function of $\kappa$, this is equivalent to $\kappa < \kappa_+$ for some $\kappa_+ > 1$.
Therefore, with these $\kappa_- < 1 < \kappa_+$, we have that for any $\kappa \in (\kappa_-, \kappa_+)$, there exists $\mathcal{L}_Y \in B_{\epsilon}(\mathcal{L}_X)$ such that $\mathbf{E} Y = \kappa \mathbf{E} X$ and so $I \subseteq  E(B_{\epsilon}(\mathcal{L}_X))$.
The cases $\mathbf{E} X \in \{0, 1\}$ are easier, so I leave them to you.

Thus, we've shown that, for any probability measure $\mathcal{L}_X \in \Delta[0, 1]$, the image of any $E(B_\epsilon(\mathcal{L}_X))$ contains an open neighbourhood of $\mathbf{E} X$, and so that $E$ is an open map.
