{\mathbb{R}}$ be an interval and consider the lower semicontinuous set-valued function $F:I\rightrightarrows {\mathbb{R}}.$ Let $J\subseteq I$ dense in $I$ and assume that $F$ is locally [strictly quasimonotone]{} on $J.$ Further, assume that $0\not\in F(x),$ for all $x\in I\setminus J.$ Then $F$ is [strictly quasimonotone]{} on $I.$

According to Lemma \[l31\], for any $x\in I$ one has $F(x)\ge 0,$ or $F(x)\le 0.$ Note that according to the hypothesis of the theorem, for any $x\in I\setminus J$ one has $F(x)>0,$ or $F(x)<0.$

We show next that $F$ is locally strictly quasimonotone on $I$, that is, every $x\in I$ admits an open neighborhood $U$, such that $F|_{U\cap I}$ is strictly quasimonotone. Let $x\in I$ and assume that $F(x)\neq\{0\}.$ Then we have $F(x)\ge 0,$ or $F(x)\le 0.$ Assume that $F(x)\ge 0$ the other case can be treated analogously. Obviously there exists $x^*\in F(x), x^*>0.$ Let $V=(x^*-\epsilon,x_1^*+\epsilon),$ an open interval, where $\epsilon>0$ such that $x^*-\epsilon>0.$ Since $F$ is lower semicontinuous, there exists an open neighborhood $U_x$ of $x$, such that for every $y\in U_x\cap I$ we have $F(y)\cap V\neq\emptyset.$ But then, according to Lemma \[l31\] $F(y)\ge 0,$ for all $y\in U_x\cap I,$ hence from Proposition \[p32\] we get that $F$ is quasimonotone on $U_x\cap I.$ 