## Upper sum equals lower sums implies constant

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

Proof so far.

Let $P = \{ a=x_1, x_2, x_3, . . . , x_{n-1},x_n=b \}$
Define $M_k = sup \{ f(x) : x \in [x_k,x_{k+1} ] \}$ and $m_k = inf \{ f(x) : x \in [x_k,x_{k+1} ] \}$

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

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

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