Let $ g(0) = a $. Then given $ \epsilon > 0 $, let $ A $ be the set where $ f(x)>a-\epsilon $, with $ |A| = 2m > 0 $. We must have then that both $ |(0,1-m)\cap A| > 0 $ and $ |(m, 1) \cap A| > 0 $. Then for $ |x| < m, g(x) \geq a - \epsilon. $
