Suppose that $\displaystyle f: [a,b] \rightarrow \mathbb {R} $ is a bounded function for which there is a partition P of $\displaystyle [a,b] $ with $\displaystyle L(f,P)=U(f,P)$. Prove that $\displaystyle f:[a,b] \rightarrow R $ is a constant.

Proof so far.

Let $\displaystyle P = \{ a=x_1, x_2, x_3, . . . , x_{n-1},x_n=b \} $

Define $\displaystyle M_k = sup \{ f(x) : x \in [x_k,x_{k+1} ] \} $ and $\displaystyle m_k = inf \{ f(x) : x \in [x_k,x_{k+1} ] \} $

Now, $\displaystyle U(f,P)= \sum ^n _{k=1} M_k(x_{k+1}-x_k) = U(f,P)= \sum ^n _{k=1} m_k(x_{k+1}-x_k) = L(f,P)$

Now, I think that $\displaystyle \sum ^n _{k=1} (x_{k+1}-x_k) = b-a$, so we have $\displaystyle M_k=m_k \ \ \ \forall k $.

So the max and min of interval are the same, would that be enough to show that f is a constant function?