An archived instance of discourse for discussion in undergraduate Real and Numerical Analysis.

Proving the Uniqueness of the Generalized Riemann Integral


Thm. 8.1.7: If a function has a generalized Riemann integral, then the value of the integral is unique.


Suppose $f$ has Generalized Riemann Integral $A_1$ and also $A_2$. Want to show that $A_1 = A_2$.

Let $\epsilon > 0$. Then there exists $\delta_1(x)$ satisfying:
$$\lvert R(f,P) - A_1 \rvert < \frac{\epsilon}{2}, \hspace{.2cm} \forall \hspace{.2cm} \delta_1(x) \text{-fine tagged partitions}$$

AND there exists $\delta_2(x)$ satisfying:
$$\lvert R(f,P) - A_2 \rvert < \frac{\epsilon}{2}, \hspace{.2cm} \forall \hspace{.2cm} \delta_2(x) \text{-fine tagged partitions}.$$
Now let $\delta(x) = min\lbrace \delta_1(x), \delta_2(x) \rbrace.$

By Thm. 8.1.5, we know there is a tagged partition $(P,\lbrace c_k \rbrace)$ that is both $\delta_1(x)$-fine and $\delta_2(x)$-fine. Then we can say $$\lvert A_1 - A_2 \rvert \leq \lvert A_1 - R(f,P) \rvert + \lvert R(f,P) - A_2 \rvert < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$
Therefore, $A_1 = A_2.\Box$


Thank you for posting all this Christy!