# Push-forward of a probability measure is mixture linear?

Let $C$ be a compact metric space, $\mathcal P(C)$ be its set of Borel probability measures (with topology of weak convergence) and $\mathcal P^2(C)$ be the set of Borel probability measures on $\mathcal P(C)$. Suppose $U:\mathcal P(C)\to \mathbb R$ is continuos. So construct $\mathcal P(U(\mathcal P(C)))$: the set of Borel probability measures on the image of $U$. So for each $p\in \mathcal P^2(C)$ I can construct the pushforward of $p$ under $U$ which is an element of $\mathcal P(U(\mathcal P(C)))$. My question is the following: the mapping $F$ from $\mathcal P^2(C)$ to $\mathcal P(U(\mathcal P(C)))$ defined as \begin{equation*} p\mapsto F(p)=p\circ U^{-1} \end{equation*}
is continuous. Is it also mixture linear? That is, if $\lambda\in[0,1]$ is $F(\lambda p+(1-\lambda)q)=\lambda F(p)+(1-\lambda)F(q)?$

Thanks in advance.

## 1 Answer

Thinking about measures over measures is a red herring here. If you reduce the problem to its essentials, it becomes very easy. If $(X,\mathcal{X})$ and $(Y,\mathcal{Y})$ are measurable spaces, $f$ a measurable function, $\mu$ and $\nu$ are probability measures on $X$, $0<\lambda<1$ and $A\in\mathcal{Y}$, then $$(\lambda\mu+(1-\lambda)\nu)\big(f^{-1}(A)\big)=\lambda \mu\big(f^{-1}(A)\big)+(1-\lambda)\nu\big(f^{-1}(A)\big).$$ The left side is the value of $A$ under the pushforward of $\lambda\mu+(1-\lambda)\nu$, and the right side is the corresponding convex combination of the value of $A$ under the pushforwards of $\mu$ and $\nu$.

• woops you're right I was overcomplicating things! Thanks! – Enrico Mar 19 '17 at 0:47