Let f, g: $Q \to R$ be bounded functions such that $f(x)\leq g(x)$, for $x \in Q$. Show that $\underline{\int_Q} f \leq \underline{\int_Q} g$ and $\overline{\int_Q} f \leq \overline{\int_Q} g$.

Note that the domain Q denotes a rectangular region may be a subset of $R^n$ for any natural number n. If the domain is just R, then Q would be a closed interval, but that shouldn't change the proof much. Also f and g are not assumed to be Riemann integrable otherwise, else the proof would be trivial.

What I have done so far is show that based on the definitions, if f and g has the same partition, then the lower sum and upper sum of f is less than or equal to the lower and upper sum of g. Even though $f\leq g$, it is possible for the upper sum of f to be greater than g if they have different partitions for some very specific partitions. So for that reason, I think we should keep the partitions the same for f and g. Now, one idea I had was to use use a sequence of partitions where we keep each subinterval (or subrectangles) the same length and just keep doubling the number of subintervals per step. At each step the upper Riemann sum of f is less than or equal to g and the limit of the partition should approach the infimum of the upper sum thus $\overline{\int_Q} f \leq \overline{\int_Q} g$. And we can do the same for the lower sum. Is this roughly correct?

  • $\begingroup$ Did you accidentally post the same question twice? $\endgroup$
    – Saad
    Jun 14 at 7:35
  • 1
    $\begingroup$ Hint: I think that there is a partition that is finer than two given partitions (one for $f$, one for $g$). $\endgroup$
    – Mindlack
    Jun 14 at 7:36
  • $\begingroup$ @Saad that is strange. I have no idea why it was posted twice. $\endgroup$
    – Bill
    Jun 14 at 7:37

The monotonicity follows directly from the definition of the lower (upper) Riemann integral as the supremum of all lower Riemann sums (resp. infimum over all upper Riemann sums) over all partitions.

For any partition $P$ is $$ L(P, f) \le L(P, g) \le \underline{\int_Q} g $$ and that implies $$ \underline{\int_Q} f = \sup_P L(P, g) \le \underline{\int_Q} g \, . $$

Similarly, $$ \overline{\int_Q} f \le U(P, f) \le U(P, g) $$ for all partitions implies that $$ \overline{\int_Q} f \le \overline{\int_Q} g \, . $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.