If $f(x):=\mu(\overline{B}(x,r))$ show that $f$ is upper semicontinuous

Question

Let $\mu$ be a Radon measure in $\mathbb R^n$, for fixed $r>0$ define $f(x):=\mu(\overline{B}(x,r))$ show that $f$ is upper semicontinuous.

Like $\overline{B}(x,r)$ is a closed set then $1_{\overline{B}(x,r)}$ is upper semicontinuous and hence given $x_n \to x$ we know that $$\limsup 1_{\overline{B}(x_n,r)} \leq 1_{\overline{B}(x,r)}$$ by the reverse Fatou´s lemma $$\limsup f(x_n)=\limsup \int1_{\overline{B}(x_n,r)} \leq\int \limsup 1_{\overline{B}(x_n,r)} \leq \int 1_{\overline{B}(x,r)}=f(x)$$ But I can't be sure if these are correct, because I can´t find a function $g \in L^{1}(\mathbb R^n ;\mu)$ such that $1_{\overline{B}(x_n,r)} \leq g$ , $\forall n \in \mathbb N$ and these is a fundamental hyp. for the reverse Fatou´s lemma.
Any hint or help? I will be very grateful.

Answer 1

Since $x_n \to x$, $x_n \in B_r(x)$ if $n > N$ for some $N$. So without loss of generality, we can assume that $x_n \in B_r(x)$ $\forall n$. Therefore, we can pick $g := \mu(B_{2r}(x))$. Since $\mu$ is a Radon measure, g should be finite.The remaining proof follows your original arguments.