Riemann Sums

Darth Frodo

Homework Statement

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.

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:

Homework Helper

Homework Statement

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.

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?

##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]##.

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

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

Oh whoops! Yes you're right.

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

Darth Frodo
No I can't say I do.

Homework Helper
No I can't say I do.

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] \}##

Darth Frodo
f(q) = 0
mi = 0

sp = ƩmiΔxi = 0

Homework Helper
f(q) = 0
mi = 0

sp = ƩmiΔxi = 0

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].

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?

Homework Helper
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?

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).

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