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I found out the proof of Urysohn's Lemma from Rudin's book but I have couple questions which I am not able to answer.

1) Why Rudin wrote that "in terms of characteristic functions, the conclusion asserts the existence of a continuous function $f$ which satisfies inequalities $\chi_K\leqslant f \leqslant \chi_V$"?

But how to conclude from here that $f$ has compact support and it lies in $V$? I concluded from $\chi_K\leqslant f \leqslant \chi_V$ that $f=1$ on $K$, $f=0$ on $V^c$. But what we can say about $f$ on $V-K$?

2) Why he mentioned semicontinuous functions? I totally didn't understand this line.

3) Each $f_r(x), g_s(x)$ are lower semicontinuous (LSC) and upper semicontinuous (USC), respectively. Right? If $r=0$ $\Rightarrow$ $f_r$ is constant function $\Rightarrow$ $f_r$ is LSC. If $r>0$ then $f_r(x)=rh_r(x)$ where $h_r$ is the indicator of open set $\Rightarrow$ $h_r$ is LSC $\Rightarrow$ $f_r$ is LSC. If $s\in \{0,1\}$ then $g_s$ is USC since it's indicator of closed or constant function, respectively. If $s\in (0,1)$ then $g_s(x)=s+(1-s)h_s(x)$ where $h_s$ is the indicator of closed set.

4) If $x\in K$ then $x\in V_r$ for all $r$ $\Rightarrow$ $f_r(x)=1$ $\Rightarrow$ $f(x)=1$. Am I right? Note that Rudin never mentioned that $K\subset V_r$ for any $r$.

5) What's motivation to show that $f=g$? I suspect that if $f=g$ then $f$ is simultaneously LSC and USC $\Rightarrow$ $f$ is continuous. Right?

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  • $\begingroup$ 1) $f$ has this property; this property itself doesn't mean $f$ is compactly supported. 2) for your (or maybe his) interest. 3) 4) 5) right. Rudin doesn't mention a lot of things. $\endgroup$
    – Ningxin
    Commented Jun 19, 2016 at 7:00
  • $\begingroup$ @QiyuWen, Your remark about 1) I didn't understand. He emphasized that exists continuous function. No words about that it's function with compact support which lies in $V$. $\endgroup$
    – RFZ
    Commented Jun 19, 2016 at 7:27
  • $\begingroup$ @QiyuWen, Maybe these properties follows from inequality $1_K\leqslant f\leqslant 1_V$? $\endgroup$
    – RFZ
    Commented Jun 19, 2016 at 7:30
  • $\begingroup$ That $f$ has compact support is because $\mathrm{supp} f \subset \bar{V}_0$. The inequalities are not enough to conclude that. $\endgroup$
    – Ningxin
    Commented Jun 19, 2016 at 7:31
  • $\begingroup$ @QiyuWen, I guess that there is some misunderstanding between our words. Sorry but look Rudin wrote that assertion of theorem is equivalent to existence of CONTINUOUS function $f$ with $1_K\leqslant f\leqslant 1_V$. Right? He doesn't mention nothing about that $f$ has compact support and that this support lies in $V$. I hope that you understood me. $\endgroup$
    – RFZ
    Commented Jun 19, 2016 at 7:44

1 Answer 1

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So $K$ and $V$ are given to us. By local compactness there exists a $U$ open satisying $K\subset U\subset \bar U\subset V$. Now apply the lemma to the pair $K,U$ to produce a (compactly supported) continuous function $f$. Since $\{x:f(x)\neq 0\}\subset U$ it follows that the support of $f$ lies in $\bar U$ and hence in $V$.

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