Using the notation from the definition of the integral, we have \(m_i \leq f(c_i) \leq M_i\text{,}\) and so

\begin{equation*}
m_i \Delta x_i \leq F(x_i) - F(x_{i-1}) \leq M_i \Delta x_i .
\end{equation*}

We sum over \(i = 1,2, \ldots, n\) to get

\begin{equation*}
\sum_{i=1}^n m_i \Delta x_i
\leq \sum_{i=1}^n \bigl(F(x_i) - F(x_{i-1}) \bigr)
\leq \sum_{i=1}^n M_i \Delta x_i .
\end{equation*}

In the middle sum, all the terms except the first and last cancel and we end up with \(F(x_n)-F(x_0) = F(b)-F(a)\text{.}\) The sums on the left and on the right are the lower and the upper sum respectively. So

\begin{equation*}
L(P,f) \leq F(b)-F(a) \leq U(P,f) .
\end{equation*}

We take the supremum of \(L(P,f)\) over all partitions \(P\) and the left inequality yields

\begin{equation*}
\underline{\int_a^b} f \leq F(b)-F(a) .
\end{equation*}

Similarly, taking the infimum of \(U(P,f)\) over all partitions \(P\) yields

\begin{equation*}
F(b)-F(a) \leq \overline{\int_a^b} f .
\end{equation*}

As \(f\) is Riemann integrable, we have

\begin{equation*}
\int_a^b f =
\underline{\int_a^b} f \leq F(b)-F(a) \leq \overline{\int_a^b} f
= \int_a^b f .
\end{equation*}

The inequalities must be equalities and we are done.