Lemma
\(f\) is Darboux integrable, then $$ \begin{align*} (U) \int_a^b f(x)dx = (L) \int_a^b f(x)dx \end{align*} $$

Proof

Since \(f\) is Darboux integrable, then by definition for all \(\epsilon > 0\), there exists a partition \(P\) such that

$$ \begin{align} U(f,P) - L(f,P) < \epsilon \quad \quad \quad (1) \end{align} $$

Recall that by definition

$$ \begin{align} (U) \int_a^b f(x)dx &= \inf\{U(f,P): P \text{ is any partition of } [a,b]\} \end{align} $$

Recall that \(U(f,P)\) is an upper Riemann sum. Among the set of all upper Riemann sums, the infimum then will satisfy

$$ \begin{align} (U) \int_a^b f(x)dx \leq U(f,P) \quad \quad \quad (2) \end{align} $$

Similarly, recall that

$$ \begin{align} (L) \int_a^b f(x)dx &= \sup\{L(f,P): P \text{ is any partition of } [a,b]\} \end{align} $$

\(L(f,P)\) is a lower Riemann sum. Among all the possible lower Riemann sums, the supremum must satisfy

$$ \begin{align} (L) \int_a^b f(x)dx \geq L(f,P) \quad \quad \quad (3) \end{align} $$

Subtract (3) from (2) to get

$$ \begin{align} (U) \int_a^b f(x)dx - (L) \int_a^b f(x)dx \leq U(f,P) - L(f,P) \end{align} $$

But from (1) we also have

$$ \begin{align} (U) \int_a^b f(x)dx - (L) \int_a^b f(x)dx \leq U(f,P) - L(f,P) < \epsilon \end{align} $$

And clearly \(U(f,P) \geq L(f,P)\) for any partition. Hence

$$ \begin{align} 0 \leq (U) \int_a^b f(x)dx - (L) \int_a^b f(x)dx \leq U(f,P) - L(f,P) < \epsilon \end{align} $$

This is true for any \(\epsilon > 0\). Hence,

$$ \begin{align} (U) \int_a^b f(x)dx = (L) \int_a^b f(x)dx \end{align} $$

References