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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 $V= B_r(x_0)=\{x|d(x,x_0)<r\}$.

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

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

How do we proceed?

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2 Answers 2

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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 $f$

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HINT

Note that

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

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

So the function is not continuous.

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