Riemann integral of a non continuous function

81 Views Asked by At

We have a function $f : I=[0,1] \rightarrow \mathbb{R}$ defined as: $$f(x)=\begin{cases} 1 &\text{if }x\in \mathbb{Q} \\ 0 &\text{if }x\in \mathbb{R}\setminus\mathbb{Q} \end{cases}$$

a) Show that for every partition V of I, we have that $\underline{S}(f,V)=0$
b) Show that $$\overline{\int}_0^1 f(x)dx =1$$

For me the difficulty with this questions is that it feels so logical but I don't know how I can really show or if I may use it that every interval on $[0,1]$ contains both rational and irrational numbers. If I some how would know/show this property I feel I can easily solve this problem. So my question is, would I have to proof this intiutive idea about intervals on $[0,1]$ for a course on analysis or is this a know result (or even "worse" is it not even true?0

1

There are 1 best solutions below

1
On BEST ANSWER

Yes, it is a fact that every nontrivial interval $(a,b)$ contains both rational and irrational numbers.

It is overwhelmingly likely that you can get away with assuming this as background knowledge.

If you do need to prove it, you need to go back to your precise assumptions and axioms about what the real numbers are. In most cases it will come down to Archimedes' axiom telling you that there is some natural number $N$ that is larger than $1/(b-a)$; the interval will then contain at least one rational with denominator $N$.

Once you have a rational $q\in(a,b)$, either $\frac{q+b}2$ or $q+\frac{b-q}{\sqrt2}$ will be an irrational between $q$ and $b$.