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Let $V \subset \mathbb R^{k}$ an open e that $\mu$ is a Borel measure positive over $\mathbb R^{k}$. The function $x \rightarrow \mu(V+x)$ is continuous? Is lower semicontinuous? Is upper semicontinuous?

Comment. If $\mu$ was Lebesgue measure we have that $m(V+x)=m(V)$ then $f: x \rightarrow \mu(V+x)$ is measurable therefore by Lusin's Lemma since $V\subset X$ and suppose $\mu(V) < \infty$ and let $\epsilon > 0$ $\exists$ compact $K \subset V$.

$\mu (V-K) < \epsilon$ and $f|_K$ is continuous.

Beside that, well I have that $\mu(E) = c m(E)$ for a constant $c \in \mathbb R^{k}$ and for all borel set $E \in \mathbb R^{k}$.

Some hint?

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    $\begingroup$ If $\mu$ is Lebesgue measure then $f\colon x\to \mu(V+x)$ is just a constant function, so it is trivially continuous. This said, try to see what happens when $V=\{0\}$ and $\mu=\delta$. $\endgroup$ Commented Jan 27, 2013 at 3:30
  • $\begingroup$ Sorry, when referring to $\mu = \delta$ it is the $\mathcal X_{E}$ measure. When $\mathcal X_{E}(x) = 1 $ if $x \in E$ and $0$ if $x \notin E$. Sorry may use different notations. $\endgroup$
    – Malaq
    Commented Jan 27, 2013 at 17:06
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    $\begingroup$ Definition of measure $\delta$ is: $\delta (E)=1$ if $0\in E$ and $\delta(E)=0$ otherwise. Don't get confused with $\chi_E$ which is a function, not a measure. $\endgroup$ Commented Jan 27, 2013 at 17:15
  • $\begingroup$ It's the Dirac Measure, right? I see in Gerald Folland Real Analysis $\endgroup$
    – Malaq
    Commented Jan 27, 2013 at 17:16
  • $\begingroup$ Yes, it is the Dirac Measure, also known as Dirac Delta. $\endgroup$ Commented Jan 27, 2013 at 17:28

2 Answers 2

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I'll give some hints since this is homework.

  1. Using @Giuseppe Negro's example with $V$ the unit ball and $\mu=\delta$ then $f(x)=1_{V}(x)$ so the only possible option is...

  2. If $x_n\to x$ and we define $g(y)=\liminf_n \chi_{x_n+V}(y)$ then $g(y)\geq \chi_{x+V}(y)$ for all $y\in \mathbb{R}^k$ (in particular $\chi_{x_n+V} \to \chi_{x+V}$ in $x+V$).

  3. Fatou's lemma.

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  • $\begingroup$ I'm thinking in the step 2 and the link to solve the problem, do not know yet. I'll think a bit more. The answer of the step 1 is upper semicontinuous. Thanks ^^. $\endgroup$
    – Malaq
    Commented Jan 27, 2013 at 18:27
  • $\begingroup$ @Malaq: Actually it should be lower semicontinuous, Giuseppe's original example gives an upper semicontinuous function, but his $V$ is closed. $\endgroup$
    – Jose27
    Commented Jan 27, 2013 at 19:59
  • $\begingroup$ When you referred to a unit ball, it was a open ball or a compact ball, because I think that was a compact ball, sorry. $\endgroup$
    – Malaq
    Commented Jan 27, 2013 at 20:25
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    $\begingroup$ @Malaq: Since your problem states $V$ open I took the open unit ball. $\endgroup$
    – Jose27
    Commented Jan 27, 2013 at 20:51
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    $\begingroup$ @Malaq: No problem, just keep that hypothesis in mind when you try to prove 2. $\endgroup$
    – Jose27
    Commented Jan 27, 2013 at 21:40
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@Jose27

I solve this exercice until one point. We tell that if $\mu$ is a Lebesgue measure, $f$ is continuous.

Take $V$ a normal ball open centered at $0$, (because is easy to see) take $\mu = \delta_{1}$, and the sequence $\{\frac{1}{k}\}$, we have that $\mu(\frac{1}{k} + V) =1$ $\forall k$. For this $f$ not be continuous, moreover $f$ not be upper semicontinuous because $\{x: f(x)<1\} = \{0\}$.

The last questions: $f$ is lower semicontinuous?

In the case above $\mu$ is.

Usually this happens?

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