 Solution

For each $x$, $g(x{)}^n\to 0$ if $g(x)<1$, and remains one if $g(x)=1$. Continuity gives pointwise convergence to $f(0)$ off $E=\{x:g(x)=1\}$ and to $f(1)$ on $E$. The functions are measurable and bounded by ${\max }_{[0,1]}|f|$, an integrable constant on a unit-length interval. The dominated convergence theorem therefore givesHere $|E|$ is Lebesgue measure. This convex combination belongs to $[f(0),f(1)]$, without requiring $f$ to be monotone.