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My question is just to make sure my proof is on the right track.

Problem: Suppose that the bounded function $f\colon [a,b]\rightarrow \mathbb{R}$ has the property that for each $x\in \mathbb{Q}$, where $a\leq x\leq b$, $f\left(x\right)=0$. Prove that

$$\underline{\int_{a}^b} f \leq 0 \leq \overline{\int_{a}^{b}} f.$$

I am going to show the part $\underline{\int_{a}^b} f \leq 0$, because the other part should be similar.

Proof:

Since $\mathbb{Q}$ is dense in $\mathbb{R}$ means that there is $x\in [x_{i-1}, x_i]$, where $f\left(x\right)=0$. Furthermore, since $$m_i\equiv \text{inf}\left\{f\left(x\right) | \; x \in [x_{i-1}, x_i]\right\}$$ means $m_i \leq 0 \: \text{for} \: i \geq 1$.

$$\implies L\left(f, P\right)= \sum_{i=1}^n m_i\left(x_i-x_{i-1}\right) \leq 0,$$ where $P$ is a partition of $[a, b]$.

Furthermore, since $$\underline{\int_{a}^b} f \equiv \: \text{sup}\left\{L\left(f, P\right)\right\},$$

$$\implies \underline{\int_{a}^b} f \leq 0.$$

Does this make sense? Thanks!

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  • $\begingroup$ is $f$ continuous ? $\endgroup$ – idm Jan 19 '15 at 18:01
  • $\begingroup$ The only stipulation is that $f$ is bounded. The problem is stated word for word. $\endgroup$ – Krampus Jan 19 '15 at 18:03
  • $\begingroup$ actually, $\int_{-a}^b f$ is not define a priori if $f:[a,b]\to \mathbb R$ $\endgroup$ – idm Jan 19 '15 at 18:04
  • $\begingroup$ Are we meant to assume that f is Reimann integrable? Certainly it must be integrable if the integral is to be defined, but a counterexample Lebesgue integrable function is easy to find. $\endgroup$ – dooglius Jan 19 '15 at 18:06
  • $\begingroup$ Yes, this is in the section for Darboux sums and integrals (6.1) problem 3 in Fitzpatrick's Advanded Calculus. $\endgroup$ – Krampus Jan 19 '15 at 18:07

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