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Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a locally bounded, discontinuous, function and let $\delta: \mathbb{R}^n \rightarrow \mathbb{R}_{\geq 0}$ be a continuous function.

Define the set-valued mapping $ F: \mathbb{R}^n \rightrightarrows \mathbb{R}^m $ as

$ F(x) := \bigcap_{r >0} \text{closure} \left( f( x+ \delta(x)\mathbb{B} + r \delta(x) \mathbb{B} ) \right), $

where $\mathbb{B}$ denotes the closed ball.

Question: is $F$ Outer SemiContinuous?

Notes.

1) It is known that the set-valued mapping $\bar{F}(x):= \bigcap_{r>0} \text{closure} \left(f(x+r \mathbb{B}) \right)$ is Outer SemiContinuous.

2) Definition of Outer SemiContinuity: a set-valued mapping $S: \mathbb{R}^n \rightrightarrows \mathbb{R}^m $ is Outer SemiContinuous at $\bar x$ if

$ \limsup_{x \rightarrow \bar x} S(x) \subset S(\bar x) $

or equivalently $\limsup_{x \rightarrow \bar x} S(x) = S(\bar x)$.

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    For continuous d: \mathbb{R}^n \rightarrow \mathbb{R}_{>0}, define G(x) := \bar{F}(x+d(x)\mathbb{B}) = \bigcap_{r>0} \text{closure} (f(x+d(x)\mathbb{B}+r\mathbb{B} )). Is $F_1$ equal to $F$ with $\delta$ strictly-positive valued?2012-09-14

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Yes. First, notice it suffices to prove that for every $y$ the set $\{x: y\notin F(x)\}$ is open. Now, if $y\notin F(x)$ then there exists $r>0$ for which the corresponding set in the intersection does not contain $y$. Finally, if $|x'-x|$ is small enough, the $r/2$-set in the definition of $F(x')$ does not contain $y$, proving the claim.

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    @Adam Yes, that is correct.2012-09-16