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Let $u,u_k \in C^{0}(K)$ where $K \subset \mathbb{R}^{n}$ is a compact set. Assume that $u_k \rightarrow u$ uniformly. Is these hypotheses sufficient to guarantee that \begin{equation} \mbox{med}(\{u_k>0\}) \rightarrow \mbox{med}(\{u>0\}) \end{equation} or \begin{equation} \mbox{med}(\{u_k \ge0\}) \rightarrow \mbox{med}(\{u \ge 0\})? \end{equation}

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What is the definition of $\mbox{med}(\{u>0\})$? – Sasha Aug 21 '12 at 19:51
This is the lebesgue measure. – user29999 Aug 21 '12 at 19:53
What is the difference between the two lines separated by 'or'? – copper.hat Aug 21 '12 at 19:53
Also, out of curiosity, where does this name for Lebesgue measure come from? I've seen many ways to denote it, but this is a first. – tomasz Aug 21 '12 at 19:54
Sorry I am going to correct. – user29999 Aug 21 '12 at 19:55
up vote 2 down vote accepted

This is not true. Let $K=[-1,1]$ and $$u_k(x)=\begin{cases} \frac{1}{k} &\text{if } x\geq \frac12\\ \frac{2x}{k} &\text{if } \frac{-1}{2}<x<\frac12\\ \frac{-1}{k} &\text{if } x\leq \frac{-1}{2}\end{cases}$$ and note that $u_k\to 0$. We have $\mathrm{med}(\{u_k>0\})=1$ for all $k$ yet $\mathrm{med}(\{u>0\})=0$, and $\mathrm{med}(\{u_k\geq 0\})=1$ for all $k$ yet $\mathrm{med}(\{u\geq 0\})=2$.

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It is not true under the given hypotheses.

Take $u_k(x) = \frac{1}{k} \max(0, \frac{1}{2}-|x-\frac{1}{2}|)$. Clearly $u_k(x) \to 0$ for all $x$, so $u = 0$. Then $m\{x | u(x) > 0 \} = 0$, whereas $m\{x | u_k(x) > 0 \} = 1$.

Furthermore, by considering $v_k = -u_k, v=-u$ instead, we have $m\{x | v(x) \geq 0 \} = 1$, whereas $m\{x | v_k(x) \geq 0 \} = 0$.

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