Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $a \in \Bbb R^*$.

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

share|cite|improve this question
"=/=" means "$\neq$"? – Michael Greinecker May 12 '12 at 18:13
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
@TonyK Is $\Bbb R^*$ good for you? – Pedro Tamaroff May 12 '12 at 18:19
Jason, I edited to add $\LaTeX$. Is everything OK? – Pedro Tamaroff May 12 '12 at 18:19
@Brian I wrote $\limsup$ first, but then flinched. Thanks. – Pedro Tamaroff May 12 '12 at 18:21

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.