Riemann Sums

1. Mar 12, 2013

Darth Frodo

1. The problem statement, all variables and given/known data
Let f(x) be deﬁned on [0,1] by

f(x) = 1 if x is rational
f(x) = 0 if x is irrational.

Is f integrable on [0,1]? You may use the fact that between any two rational numbers
there exists an irrational number, and between any two irrational numbers there exists
a rational number.

2. Relevant equations

3. The attempt at a solution

Divide into n sub-intervals.

Δxi=1/n

U(f,Pn) = Ʃ(f(Ui)Δxi) = $\sum(1)(1/n)$ = 1/n
L(f,Pn) = Ʃ[f(li)](Δxi) = $\sum(0)(1/n)$ = 0

As n$\rightarrow$ $\infty$ both U(f,Pn) and L(f,Pn) $\rightarrow$ 0

Therefore $\int f(x)dx$ = 0

Is this correct?

Last edited: Mar 12, 2013
2. Mar 12, 2013

Zondrina

$f$ is most certainly not integrable on any interval $[0,b]$.

For each sub interval $[x_{i-1}, x_{i}]$ you can always find a rational number $r \in [x_{i-1}, x_{i}]$ with $f(r) = 0$.

So this tells you something about the lower sum.

We can also find an irrational number $q \in [x_{i-1}, x_{i}]$ such that $f(q) = 1$.

This tells you something about the upper sum.

Using this you can deduce $f$ is not integrable on any interval $[0,b]$.

3. Mar 12, 2013

Darth Frodo

Shouldn't it be f(r) = 1 and f(q) = 0?

4. Mar 12, 2013

Zondrina

Oh whoops! Yes you're right.

My bad about that, but you see what I was getting at right?

5. Mar 12, 2013

Darth Frodo

No I can't say I do.

6. Mar 12, 2013

Zondrina

Well... lets do the upper sum then as an example.

Since $f(r) = 1$, you know $M_i = 1$ no matter what sub-interval you choose.

The upper sum is defined as such :

$S_p = \sum M_i Δx_i = \sum 1 Δx_i = b$

Now try doing the lower sum $s_p$, what do you get?

EDIT : Just for clearness $M_i = sup\{ f(x) \space | \space x \in [x_{i-1}, x_i] \}$

7. Mar 12, 2013

Darth Frodo

f(q) = 0
mi = 0

sp = ƩmiΔxi = 0

8. Mar 12, 2013

Zondrina

Exactly. So sup(sp) = 0 and inf(Sp) = b.

Since 0 ≠ b, we know that f can't be Riemann integrable on any [0,b].

You can actually generalize it to any [a,b].

9. Mar 12, 2013

Darth Frodo

Ok, Thank you very much. But would you mind telling me what I did wrong in the 1st post? What assumption of mine is incorrect?

10. Mar 12, 2013

Zondrina

You took n to infinity which wasn't what you were supposed to do.

A function is Riemann Integrable on an interval [a,b] if sup(sp) = inf(Sp).

11. Mar 13, 2013

SammyS

Staff Emeritus
$\displaystyle U(f,P_n)=\sum_{i=1}^{n}(1)(1/n)=n(1/n)=1$