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}
A natural question about convergence
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analysis
measure-theory
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0What is the definition of $\mbox{med}(\{u>0\})$? – 2012-08-21
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0This is the lebesgue measure. – 2012-08-21
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0What is the difference between the two lines separated by 'or'? – 2012-08-21
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4Also, 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. – 2012-08-21
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0Sorry I am going to correct. – 2012-08-21
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0Try it out with $u=0$ and $u_k>0$ or $u_k<0$ respectively. – 2012-08-21
2 Answers
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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}
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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 $m\{x | v_k(x) \geq 0 \} = 0$.