# Is f continuous? lower semicontinuous? upper semicontinuous?

I'm learning analysis on my own, and I'm having difficulties.
Let $$\mu_{x_0}(E)=\begin{cases} 1 & x_0\in E \\ 0 & x_0\not\in E \end{cases}$$ and $$V= B_r(x_0)=\{x|d(x,x_0).

Is $$f(x)=\mu(V+x)$$ continuous? lower semicontinuous? upper semicontinuous?

My attempt:
So $$V+x=B_r(x_0+x)=\{x|d(x,x_0+x)
$$d(x,x_0+x) \leq d(x,x)+d(x_0,x) < r$$
so $$x_0 \not\in V+x,$$ so $$\mu(V+x)=0=f(x)$$

How do we proceed?

As noted in another answer, $$f$$ is the indicator function of an open set (the open ball centered at $$x_0$$ of radius $$r$$). As such, $$f$$ is lower semi-continuous.

Exercise: Let $$g(x):=[r-|x-x_0|]^+$$ be the distance from $$x$$ to the complement of $$B_r(x_0)$$. Clearly $$g$$ is continuous. Show that as $$n\to\infty$$, the sequence $$\{f_n\}$$ of continuous functions defined by $$f_n(x) :=g(x)/[n^{-1}+g(x)]$$ increases pointwise to $$f$$

HINT

Note that

$$\mu(V+x)=1$$ if $$d(x,x_0) and

$$\mu(V+x)=0$$ if $$d(x,x_0) \geq r$$

So the function is not continuous.