# Closure of Locally-Bounded function is Outer SemiContinuous?

Consider a locally-bounded set-valued mapping $f: \mathbb{R}^n \rightrightarrows \mathbb{R}^m$ and the set-valued mapping $F: \mathbb{R}^n \rightrightarrows \mathbb{R}^m$ defined as

$$F(x) := \text{closure}(f(x)).$$

Question: is the mapping $F$ Outer SemiContinuous?

Note: definition of Outer SemiContinuity for a set-valued map.

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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The definition of $F$ is not clear. What do you mean? – Jochen Sep 7 '12 at 12:17
Did you mean $f\colon \mathbb{R}^n \rightrightarrows \mathbb{R}^m$? Otherwise $\mathop{\mathrm{closure}} f(x)$ doesn't seem to make sense. – martini Sep 7 '12 at 12:51
Martini, you are right. I meant $f$ set-valued. Thanks. – Adam Sep 7 '12 at 13:58

Counterexample in one dimension: $f(x)=0$ if $|x|\le1$ and $f(x)=1$ otherwise. Here $F=f$, which is not outer semicontinuous.