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Let $$I_{u}(x) = \begin{cases} 1, & \quad x \geq 0 \\ 0, & \quad x < 0 \end{cases}$$ and $$I_{l}(x) = \begin{cases} 1, & \quad x > 0 \\ 0, & \quad x \leq 0. \end{cases}$$ Evidently, $I_{u}(x)$ is upper semicontinuous whereas $I_{l}(x)$ is lower semicontinuous.

Let $L(x,\nu)$ be such that, $\forall x \in \mathbb{R}$, $\lim_{\nu \to \infty} L(x,\nu) = I_{u}(x)$: this means that, for any fixed $\epsilon>0$, $\exists \bar{\nu}$ such that, $\forall x \in \mathbb{R}$, it holds that $\big| L(x,\nu) - I_{u}(x) \big| \leq \epsilon$, $\forall \nu \geq \bar{\nu}$.

Then, can we also say that $\lim_{\nu \to \infty} L(x,\nu) = I_{l}(x)$?

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1 Answer 1

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Hint: examine what happens at $x=0$.

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  • $\begingroup$ I've just realized it's a stupid question: the answer is obviously NO. $\endgroup$
    – TheDon
    May 26, 2014 at 18:46

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