Proof concerning the Riemann integral

Question

Let $f$ be continuous on [0,1] and suppose that $1 \ge f(x) \ge 0$ for all $x \in [0,1]$. Prove that if the upper integral $U(f) = 1$, then $f(x) = 1$ for all $x \in [0,1]$

I feel like this is really easy. Can you someone help me?

Answer 1

Assume $f(x_0)<1$ for some point $x_0\in[0,1]$. Pick $q$ with $f(x_0)<q<1$. Show that there is an interval $[u,v]\subset [0,1]$ with $f(x)\le q$ for $x\in[u,v]$. Show that $U(f)\le1-(1-q)(b-a)$.