Is $\mu_U(E) := \mu(E\cap U)$ a Radon measure when $U$ is open?

Question

Let $X$ be a LCH space with a Radon measure $\mu$. By definition, Radon measure is a Borel measure, which is finite on compact sets, outer regular on Borel sets, and inner regular on open sets.
Now, let $U \subseteq X$ be open. I want to prove/disprove that $\mu_U(E) := \mu(E\cap U)$ is also Radon. It is easy to prove that $\mu_U$ is finite on compact sets and inner regular on open sets.
At the moment, I am not sure whether this is true/false. I would appreciate if you can help me with proving/disproving this.

Answer 1

Unfortunately it isn't a Radon measure in general. However, there is a Radon measure which agrees with $\mu_U$ when integrating compactly supported continuous functions (by Riesz representation theorem).

It is possible to remedy this though by using a different definition of Radon. If Radon instead means finite on compact sets and inner regular on Borel sets, then $\mu_E$ is Radon for every Borel set $E$.