[proofplan]
We interpret $\mathbb P\circ X^{-1}$ as the pushforward set function $A\mapsto \mathbb P(X^{-1}(A))$ on $\mathcal E$. Measurability of $X$ ensures that every inverse image $X^{-1}(A)$ with $A\in\mathcal E$ belongs to $\mathcal F$, so the expression is well-defined. The measure axioms follow by transferring the empty set, disjointness, and countable unions through inverse images, then applying the corresponding axioms for $\mathbb P$. Finally, the total mass is one because $X^{-1}(E)=\Omega$.
[/proofplan]
[step:Define the pushforward set function and check that it is well-defined]
Define $\mu:\mathcal E\to[0,1]$ by
\begin{align*}
\mu(A)=\mathbb P(X^{-1}(A))
\end{align*}
for every $A\in\mathcal E$. Since $X:\Omega\to E$ is $(\mathcal F,\mathcal E)$-measurable, $X^{-1}(A)\in\mathcal F$ for every $A\in\mathcal E$. Hence $\mathbb P(X^{-1}(A))$ is defined. Since $\mathbb P$ is a probability measure, $0\leq \mathbb P(B)\leq 1$ for every $B\in\mathcal F$, and therefore $\mu(A)\in[0,1]$ for every $A\in\mathcal E$.
Also,
\begin{align*}
\mu(\varnothing)=\mathbb P(X^{-1}(\varnothing))=\mathbb P(\varnothing)=0.
\end{align*}
Thus $\mu$ is a well-defined nonnegative set function on $\mathcal E$ with empty-set value zero.
[/step]
[step:Transfer countable disjoint unions through inverse images]
Let $(A_n)_{n\in\mathbb N}$ be a sequence of pairwise disjoint sets in $\mathcal E$. For each $n\in\mathbb N$, define $B_n:=X^{-1}(A_n)$. By measurability of $X$, each $B_n$ belongs to $\mathcal F$.
First, the sets $(B_n)_{n\in\mathbb N}$ are pairwise disjoint. Indeed, if $m,n\in\mathbb N$ with $m\neq n$, then
\begin{align*}
B_m\cap B_n=X^{-1}(A_m)\cap X^{-1}(A_n)=X^{-1}(A_m\cap A_n)=X^{-1}(\varnothing)=\varnothing.
\end{align*}
Second, inverse images commute with arbitrary unions, so
\begin{align*}
X^{-1}\left(\bigcup_{n=1}^{\infty}A_n\right)=\bigcup_{n=1}^{\infty}X^{-1}(A_n)=\bigcup_{n=1}^{\infty}B_n.
\end{align*}
[guided]
We need to prove countable additivity of $\mu$, and $\mu$ is defined by first taking an inverse image and then applying $\mathbb P$. Therefore the central question is: what happens to a disjoint countable union under $X^{-1}$?
Let $(A_n)_{n\in\mathbb N}$ be pairwise disjoint in $\mathcal E$, and define $B_n:=X^{-1}(A_n)$ for each $n\in\mathbb N$. The measurability of $X:\Omega\to E$ is exactly what lets us pass from the target $\sigma$-algebra $\mathcal E$ back to the source $\sigma$-algebra $\mathcal F$: since $A_n\in\mathcal E$, we have $B_n\in\mathcal F$. Thus the probabilities $\mathbb P(B_n)$ are all meaningful.
We next verify disjointness. If $m\neq n$, then $A_m\cap A_n=\varnothing$ because the original sequence is pairwise disjoint. For any $\omega\in\Omega$, the condition $\omega\in X^{-1}(A_m)\cap X^{-1}(A_n)$ means both $X(\omega)\in A_m$ and $X(\omega)\in A_n$, equivalently $X(\omega)\in A_m\cap A_n$. Hence
\begin{align*}
X^{-1}(A_m)\cap X^{-1}(A_n)=X^{-1}(A_m\cap A_n)=X^{-1}(\varnothing)=\varnothing.
\end{align*}
So the sequence $(B_n)_{n\in\mathbb N}$ is pairwise disjoint in $\mathcal F$.
Finally, we verify the countable-union identity. For $\omega\in\Omega$, membership in the inverse image of the union means
\begin{align*}
\omega\in X^{-1}\left(\bigcup_{n=1}^{\infty}A_n\right)
\end{align*}
if and only if
\begin{align*}
X(\omega)\in\bigcup_{n=1}^{\infty}A_n.
\end{align*}
This holds if and only if there exists $n\in\mathbb N$ such that $X(\omega)\in A_n$, which is equivalent to the existence of $n\in\mathbb N$ such that $\omega\in X^{-1}(A_n)$. Therefore
\begin{align*}
X^{-1}\left(\bigcup_{n=1}^{\infty}A_n\right)=\bigcup_{n=1}^{\infty}X^{-1}(A_n)=\bigcup_{n=1}^{\infty}B_n.
\end{align*}
This identity is the bridge that allows countable additivity of $\mathbb P$ on $\mathcal F$ to become countable additivity of $\mu$ on $\mathcal E$.
[/guided]
[/step]
[step:Apply countable additivity of $\mathbb P$ to obtain countable additivity of $\mu$]
Let $(A_n)_{n\in\mathbb N}$ be a sequence of pairwise disjoint sets in $\mathcal E$, and define $B_n:=X^{-1}(A_n)$ for each $n\in\mathbb N$. From the previous step, $(B_n)_{n\in\mathbb N}$ is pairwise disjoint in $\mathcal F$ and
\begin{align*}
X^{-1}\left(\bigcup_{n=1}^{\infty}A_n\right)=\bigcup_{n=1}^{\infty}B_n.
\end{align*}
Using the definition of $\mu$ and countable additivity of the measure $\mathbb P$ on $\mathcal F$, we obtain
\begin{align*}
\mu\left(\bigcup_{n=1}^{\infty}A_n\right)=\mathbb P\left(X^{-1}\left(\bigcup_{n=1}^{\infty}A_n\right)\right).
\end{align*}
Thus
\begin{align*}
\mu\left(\bigcup_{n=1}^{\infty}A_n\right)=\mathbb P\left(\bigcup_{n=1}^{\infty}B_n\right)=\sum_{n=1}^{\infty}\mathbb P(B_n).
\end{align*}
Since $B_n=X^{-1}(A_n)$, this becomes
\begin{align*}
\mu\left(\bigcup_{n=1}^{\infty}A_n\right)=\sum_{n=1}^{\infty}\mathbb P(X^{-1}(A_n))=\sum_{n=1}^{\infty}\mu(A_n).
\end{align*}
Therefore $\mu$ is countably additive on $\mathcal E$.
[/step]
[step:Compute the total mass]
Since $X:\Omega\to E$, every $\omega\in\Omega$ satisfies $X(\omega)\in E$. Hence $X^{-1}(E)=\Omega$. Therefore
\begin{align*}
\mu(E)=\mathbb P(X^{-1}(E))=\mathbb P(\Omega)=1.
\end{align*}
The last equality holds because $(\Omega,\mathcal F,\mathbb P)$ is a [probability space](/page/Probability%20Space).
We have shown that $\mu$ is a countably additive nonnegative set function on $\mathcal E$, that $\mu(\varnothing)=0$, and that $\mu(E)=1$. Hence $\mu$ is a probability measure on $(E,\mathcal E)$. Since $\mu=\mathbb P\circ X^{-1}$ by definition, $\mathbb P\circ X^{-1}$ is a probability measure on $(E,\mathcal E)$.
[/step]
