Riemann Sums

  • #1
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1

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.

Δxi=1/n

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

As n[itex]\rightarrow[/itex] [itex]\infty[/itex] both U(f,Pn) and L(f,Pn) [itex]\rightarrow[/itex] 0

Therefore [itex]\int f(x)dx[/itex] = 0

Is this correct?
 
Last edited:

Answers and Replies

  • #2
STEMucator
Homework Helper
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140

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.

Δxi=1/n

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

As n[itex]\rightarrow[/itex] [itex]\infty[/itex] both U(f,Pn) and L(f,Pn) [itex]\rightarrow[/itex] 0

Therefore [itex]\int f(x)dx[/itex] = 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]##.
 
  • #3
210
1
Shouldn't it be f(r) = 1 and f(q) = 0?
 
  • #4
STEMucator
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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?
 
  • #6
STEMucator
Homework Helper
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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] \}##
 
  • #8
STEMucator
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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].
 
  • #9
210
1
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
STEMucator
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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).
 
  • #11
SammyS
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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?

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

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