I really could use a hand on this question
Let $f:\mathbb{R} \mapsto f(\mathbb{R}):=E $, a continuous function. Let $\mu_e$ be a probability on $(E,B(E))$.
Show that there exists a measure $\nu$ on $\mathbb{R}$ such that :
$\forall B\in B(E), \mu_e(B) = \nu(f^{-1}(B))$
I really can't think of a single way to approache this. The obvious choice would seem to be $\nu(B) = \mu_e(f(B))$, but this doesn't seem to work. Another idea would be to construct an outer measure with one of the methods available but I don't know how I would go about it
I thank you all in advance
Hint:
$E$ is an interval (possibly a single point) since $f$ is continuous. Let $\mathcal{A}=\{f^{-1}(B): B\in\mathscr{B}(E)\}$. This is a $\sigma$-algebra, define $\nu(f^{-1}(B))=\mu(B)$.
$\nu$ is well defined since $f:\mathbb{R}\rightarrow E$ is surjective; thus, if $f^{-1}(B)=f^{-1}(B')$, then $B=f(f^{-1}(B))=f(f^{-1}(B'))=B'$.
If $\{f^{-1}(B_n):n\in\mathbb{N}\}$ are disjoint, then $\{B_n:n\in\mathbb{N}\}$ are disjoint since $f$ is surjective. Thus, if the $\{B_n:n\in\mathbb{N}\}\subset\mathscr{B}(E)$, $$\nu\big(\bigcup_nf^{-1}(B_n)\big)=\nu\big(f^{-1}(\bigcup_nB_n)\big)=\mu(\bigcup_nB_n)=\sum_n\mu(B_n)=\sum_n\nu(f^{-1}(B_n))$$