Integrability of f(x) on [0,1]

[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.

Homework Equations

The Attempt at a Solution

Divide into n sub-intervals.

Δx_i=1/n

U(f,P_n) = Ʃ(f(U_i)Δx_i) = \sum(1)(1/n) = 1/n
L(f,P_n) = Ʃ[f(l_i)](Δx_i) = \sum(0)(1/n) = 0

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

Therefore \int f(x)dx = 0

Is this correct?

 

[STEMucator]

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.

Homework Equations

The Attempt at a Solution

Divide into n sub-intervals.

Δx_i=1/n

U(f,P_n) = Ʃ(f(U_i)Δx_i) = \sum(1)(1/n) = 1/n
L(f,P_n) = Ʃ[f(l_i)](Δx_i) = \sum(0)(1/n) = 0

As n\rightarrow \infty both U(f,P_n) and L(f,P_n) \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?

 

[STEMucator]

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.

 

[STEMucator]

No I can't say I do.

Well... let's 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
m_i = 0

s_p = Ʃm_iΔx_i = 0

 

[STEMucator]

f(q) = 0
m_i = 0

s_p = Ʃm_iΔx_i = 0

Exactly. So sup(s_p) = 0 and inf(S_p) = 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?

 

[STEMucator]

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(s_p) = inf(S_p).

 

[SammyS]

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?

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