Semi continuity and Borel Sets - Measurable Functions

[SqueeSpleen]

In a book I'm reading it says:
\newline
If $f: \mathbb{R} \longrightarrow \mathbb{R}$ is lower semi continous, then $\{f > a \}$ is an open set therefore a borel set. Then all lower semi continuous functions are borel functions.
It's stated as an obvious thing but I couldn't prove it.
The definition a lower semi continuous function I'm using is:
A function is lower semicontinous in the point $x_0$ if
$\underline{lim}_{x \rightarrow x_{0}} \geq f(x_{0})$
$\underline{lim}_{x \rightarrow x_{0}}= \sup_{\delta > 0 } \{ inf \{ f(x) / |x-x_{0} | < \delta \} \}$
(A function is lower semicontinous in $\mathbb{R}^{p}$ if for all $x \in \mathbb{R}^{p}$ it's lower semicontinous).
[itex][/itex]
Can someone give me a hint please?

 

[lippyka]

I'd really appreciate it. The key idea here is that if a function is lower semi-continuous, then its inverse image with respect to any real number is an open set, and hence a Borel set. More specifically, let $f:\mathbb{R}\to\mathbb{R}$ be a lower semi-continuous function, and let $a\in\mathbb{R}$. Then the set $\{x\in\mathbb{R}:f(x)>a\}$ is open. To see why, suppose $x_0\in\{x\in\mathbb{R}:f(x)>a\}$. Then by the definition of lower semi-continuity, there exists some $\delta>0$ such that for all $x$ with $|x-x_0|<\delta$, we have $f(x)\geq f(x_0)>a$. This means that the set $\{x\in\mathbb{R}:f(x)>a\}$ is a union of open intervals centered at each point in the set, and hence is open itself. Since this set is open, it is also a Borel set, and thus all lower semi-continuous functions are Borel functions.