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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.

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    "=/=" means "$\neq$"?2012-05-12
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    The range of $f$ has to be $\mathbb R \cup \{-\infty,+\infty\}$ for your definition to make sense.2012-05-12
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    @TonyK Is $\Bbb R^*$ good for you?2012-05-12
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    Jason, I edited to add $\LaTeX$. Is everything OK?2012-05-12
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    @Brian I wrote $\limsup$ first, but then flinched. Thanks.2012-05-12
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    @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 $+\infty$?2012-05-12
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    @TonyK I think I saw it in Rudin's or Apostol's Analysis, to symbolize the extended reals.2012-05-12
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    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](http://math.stackexchange.com/questions/123379/show-that-if-f-1-alpha-infty-is-open-for-any-alpha-in-mathbbr)2012-05-12
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    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 $\mathbb R$.2012-05-12
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    @TonyK: I doubt it; I'm just not quite awake yet. It seems likely that $f$ is upper semicont. everywhere.2012-05-12

1 Answers 1

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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 $L=\{x:f(x)0$. 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)

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

$$E=\bigcap_{n\in\Bbb N}\{x:f(x)

and apply the first part of the problem.