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$M$ is a compact space. Assume $f$ is upper semi-continuous on $M$, $g$ is lower semi-continuous on $M$, and $f(x) \geq g(x)$ for any $x\in M$.

If $f(x_0) = g(x_0) $ for some point $x_0\in M$, is it true that $f$ is continuous at $x_0$? (I think it is true just using the definition of semi-continuous functions using $\limsup_{x\rightarrow y} f(x) \leq f(y)$ and $\liminf_{x\rightarrow y} g(x) \geq g(y)$)

What about the topology of the set of the continuous points of $f$? Or the topology of the set of $x$ such that $f(x) = g(x)$ (open or closed?)

Thank you!

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closed as off-topic by Pietro Majer, Wolfgang, Alexey Ustinov, Alex Degtyarev, Stefan Kohl Dec 1 '16 at 13:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Pietro Majer, Wolfgang, Alexey Ustinov, Alex Degtyarev, Stefan Kohl
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ What if $X=[-1,1]$, $f_n$ are all equal to the characteristic function of $[0,1]$, $g_n$ to that of $(0,1]$? $\endgroup$ – Fedor Petrov Nov 30 '16 at 15:00
  • $\begingroup$ Thank Fedor Petrov! What about $\{f_n\}_{n\geq1}$ is lower semi continuous, decreasing and $\{g_n\}_{n\geq1}$ is upper semi-continuous, increasing and $f_n(x)\geq g_n(x)$? Is it true to construct a continuous function $h(x)$ in between? $\endgroup$ – Xifeng Su Nov 30 '16 at 19:00
  • $\begingroup$ Xifeng Su: no, not even if you assume all $f_n$ and $g_n$ to be continuous: take $X:=[0,1]$, $g_n(x)=\min( nx_+,1)$, $f_n(x)=\min( (nx+1)_+,1)$. $\endgroup$ – Pietro Majer Nov 30 '16 at 21:56
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    $\begingroup$ It is true that for any $f$ LSC and $g$ USC on $X$, with $f\ge g$, there is a continuous $h$ in between. It's Katetov's insertion theorem, actually a characterization of normal spaces $X$. $\endgroup$ – Pietro Majer Nov 30 '16 at 22:00
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In general there is no continuous function between u.s.c. $f$ and l.s.c. $g\leq f$. For example, take $f(x)=1, 0\leq x\leq 1;\; f(x)=0, 1<x\leq 2$, this is u.s.c. Now $g(x)=1, 0\leq x<1;\; g(x)=0, 1\leq x\leq 2$, this is l.c.s, and $g(x)\leq f(x)$ and evidently there is no continuous function in between.

Moreover, it is easy to arrange a decreasing sequence of continuous function tending to $f$ from above, and increasing sequence of continuous functions tending to $g$ from below. This the answer to your question is no, even if your functions $f_n$, $g_n$ are continuous.

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