Claim: For all $ a \in \mathbb{R}, A_{0} = \{ x \in \mathbb{R} : g(x) > a \} $ is open relative to $ \mathbb{R} $.
Proof: Suppose $ \exists x \in A_{0} $, fix this x. Then $ g(x) > a $, and
$ \exists $ a set $ E \subseteq (x,x+1) $ such that $ f(y) > a $ on E, and $ |E| > 0 $. Suppose $ |E| = \gamma > 0 $.
I want to show that there is a small ball of
radius $ \delta $ in $ \mathbb{R} $ such that this small open ball is also in $ A_{0} $.
Choose $ \delta > 0 $ like this: take $ \delta = \gamma/2 $.
Then, $ \forall y \in \mathbb{R} $ such that $ |x - y| < \delta $, we have
$ | (y,y+1) \bigcap E | \geq \delta = \gamma/2 > 0 $. Therefore, $ g(y) \geq g(x) > a. $
And thus, for all $ y \in \mathbb{R} $ such that $ |x-y| < \delta, y \in A_{0} $ as well!
So, $A_{0}$ is open relative to $R$, and Theorem (4.14) on page 56 tells
us that g is lsc at x=0 (i.e. if a = g(0)).
