Pushforward measure is Borel regular

39 Views Asked by At

Let $\mu$ be an outer measure in $\mathbb R^n$ and $f:\mathbb R^n \to \mathbb R^m$ a function i know that if $\mu$ is Radon and $f$ continuous and proper then the pushforward measure $f_{\sharp}(\mu)$ is Radon, my question is: Why under conditions of $\mu$ and $f$ i can conclude that $f_{\sharp}(\mu)$ is Borel regular?

Any hint or help o will be very grateful, im interesting to some reference of examples