Prove that $\{ x: f(x) < a \}$ is an open set. Prove that $\{ x: f(x) = a \}$ is a Borel set.

Question

A function $f:\Bbb R\to\Bbb R^*$ ($\Bbb R^*$ is the reals together with $\pm\infty$) is upper semicontinuous at $y$ if

$f(y)\neq +\infty$ and $f(y) \geq \limsup\limits_{x\to y} f(x)$. Let $a \in \Bbb R^*$.

The range of $f$ has to be $\mathbb R \cup {-\infty,+\infty}$ for your definition to make sense. — TonyK, May 12 '12 at 18:18
@Peter: I'm not familiar with the $\mathbb R^*$ notation. But in any case the function is described as "f: R->R" in the first line. So how could $f(y)$ ever equal $+\infty$? — TonyK, May 12 '12 at 18:23
@TonyK I think I saw it in Rudin's or Apostol's Analysis, to symbolize the extended reals. — Pedro, May 12 '12 at 18:24
Closely related question - but it uses a different (but probably equivalent) definition of semicontinuity: Show that if $f^{-1}((\alpha, \infty))$ is open for any $\alpha \in \mathbb{R}$, then $f$ is lower-semicontinuous — Martin Sleziak, May 12 '12 at 18:29
The question is still incomplete. Is $f$ supposed to be upper semi-continuous everwhere? The question doesn't say so. Also, is the domain of $f$ really $\mathbb R^*$, as per Brian's edit? Normally it would just be $\mathbb R$. — TonyK, May 12 '12 at 18:31
@TonyK: I doubt it; I'm just not quite awake yet. It seems likely that $f$ is upper semicont. everywhere. — Brian M. Scott, May 12 '12 at 18:36

Brian M. Scott · Answer 1 · 2012-05-12T20:51:17.237

Suppose that $f$ is upper semicontinous; then for each $x_0\in\Bbb R$ and $\epsilon>0$ there is a $\delta>0$ such that $f(x)\le f(x_0)+\epsilon$ whenever $|x-x_0|<\delta$. Fix $a\in\Bbb R^*$, and let $L=\{x:f(x)<a\}$; we wish to show that $L$ is open. Let $x_0\in L$ be arbitrary. Let $\epsilon=\frac12(a-f(x_0))$; $x_0\in L$, so $f(x_0)<a$, and $\epsilon>0$. Can you finish the argument from there? I've completed it but left it spoiler-protected.

By hypothesis there is a $\delta>0$ such that $f(x)\le f(x_0)+\epsilon$ whenever $|x-x_0|<\delta$. But $$f(x_0)+\epsilon=f(x_0)+\frac12\Big(a-f(x_0)\Big)=\frac12\Big(f(x_0)+a\Big)<a\;,$$ so $f(x)<a$ whenever $|x-x_0|<\delta$, and therefore $(x_0-\delta,x_0+\delta)$ is an open interval around $x_0$ contained in $L$. Since $x_0$ was an arbitrary point of $L$, $L$ is open.

For the second result, let $E=\{x:f(x)=a\}$. Observe that

$$E=\bigcap_{n\in\Bbb N}\{x:f(x)<a+2^{-n}\}\cap\Big(\Bbb R\setminus\{x:f(x)<a\}\Big)\;,$$

and apply the first part of the problem.

Prove that $\{ x: f(x) < a \}$ is an open set. Prove that $\{ x: f(x) = a \}$ is a Borel set.

1 Answers1