Show [itex]\phi[/itex][itex]\circ[/itex]f is Riemann integrable

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In summary, the homework statement is that if a function is Riemann integrable, then there exists a partition P such that U(P, f) - L(P, f) < \epsilon.
  • #1
IntroAnalysis
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Homework Statement



Let f:[a, b][itex]\rightarrow[/itex][m, M] be a Riemann integrable function and let
[itex]\phi[/itex]:[m, M][itex]\rightarrow[/itex]R be a continuously differentable function
such that [itex]\phi[/itex]'(t) [itex]\geq[/itex]0 [itex]\forall[/itex]t (i.e. [itex]\phi[/itex]
is monotone increasing). Using only Reimann lemma, show that the composition [itex]\phi[/itex][itex]\circ[/itex]f is Riemann integrable.

Homework Equations


Riemann lemma - f: [a, b] [itex]\rightarrow[/itex] is Riemann integrable iff for any [itex]\epsilon[/itex]>0 [itex]\exists[/itex]a partition P such that U(P, f) - L(P, f) < [itex]\epsilon[/itex].

Function f is Riemann integrable hence it is bounded by [m, M]. Thus [itex]\forall[/itex]
x[itex]\in[/itex][a, b],lf(x)l [itex]\leq[/itex] max{m, M}.

Also, since the domain of [itex]\phi[/itex] is compact and the function is monotone and increasing, by the Extreme Value Theorem, it achieves a maximum and a minimum on [m, M], hence [itex]\phi[/itex] is also bounded. Thus, [itex]\phi[/itex]((f(a)) and [itex]\phi[/itex](f(b)) is bounded by some constant, K.


Also know since f is Riemann integrable that there exists a partition P such that
U(P, f) - L (P, f)< [itex]\epsilon[/itex]

We must show U(P,[itex]\phi[/itex](f(x))) - L(P, [itex]\phi[/itex](f(x)))<[itex]\epsilon[/itex].

I think I have most of the major pieces, can someone suggest how to put it together?
Thank you.
 
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  • #2
Why don't you start by writing out the definition of

[tex]U(P,f)-L(P,f)[/tex]
 
  • #3
The upper Riemann sum = (i=1[itex]\rightarrow[/itex]n) [itex]\sum[/itex]MiΔxi where Δxi=[xi-xi-1]

The lower Riemann sum = i=(1[itex]\rightarrow[/itex]n) [itex]\sum[/itex]miΔxi where Δxi=[xi-xi-1]

Mi=sup[f(xi): x[itex]\in[/itex][xi, xi-1]
mi=inf[f(xi): x[itex]\in[/itex][xi, xi-1]
 
  • #4
OK, let [itex]c_i[/itex] (resp. [itex]d_i[/itex]) be the element of [itex][x_i,x_{i+1}][/itex], where f reaches his maximum (resp. minimum).

We know that

[tex]\sum_{i=1}^n (f(c_i)-f(d_i)) \Delta x_i<\varepsilon[/tex]

Now, can you prove that [itex]\varphi\circ f[/itex] also reaches his maximum (resp. minimum) in [itex]c_i[/itex] (resp. [itex]d_i[/itex])??

If that were true, then we have to do something with


[tex]\sum_{i=1}^n (\varphi (f(c_i))-\varphi(f(d_i))) \Delta x_i<\varepsilon[/tex]
 
  • #5
Can't we just say that since [itex]\phi[/itex] is monotone increasing, that we know that [itex]\phi[/itex](f(ci)) (resp. [itex]\phi[/itex](f(di))) is where [itex]\phi[/itex] reaches its maximum (resp. minimum)?

Thus, 0[itex]\leq[/itex]l [itex]\phi[/itex](f(ci) - [itex]\phi[/itex](f(di)) l [itex]\leq[/itex]2K?
 
  • #6
IntroAnalysis said:
Thus, 0[itex]\leq[/itex]l [itex]\phi[/itex](f(ci) - [itex]\phi[/itex](f(di)) l [itex]\leq[/itex]2K?

I don't see where this comes from or why it is necessary.

But, basically, we have something of the form

[tex]\sum_{i=1}^n {(\varphi(f(c_i))-\varphi(f(d_i)))\Delta x_i}[/tex]

and you must associate this with

[tex]\sum_{i=1}^n {(f(c_i)-f(d_i))\Delta x_i}[/tex]

Do you have any result that associates [itex]\varphi(f(c_i))-\varphi(f(d_i))[/itex] with [itex]f(c_i)-f(d_i)[/itex]?? (use that [itex]\varphi[/itex] is differentiable)
 

FAQ: Show [itex]\phi[/itex][itex]\circ[/itex]f is Riemann integrable

1. What is the definition of Riemann integrability?

The Riemann integral, also known as the Riemann sum, is a method for calculating the area under a curve by dividing the region into smaller, simpler shapes and summing their areas. A function is considered Riemann integrable if the limit of these sums approaches a finite value as the size of the shapes approaches zero.

2. How do you show that a function is Riemann integrable?

To show that a function is Riemann integrable, one must demonstrate that the limit of the Riemann sums exists and is equal to a finite value. This can be done through various techniques, such as the Riemann sum definition, the Darboux integral definition, or the Cauchy criterion.

3. Can a function be Riemann integrable on a closed interval but not on an open interval?

Yes, a function can be Riemann integrable on a closed interval but not on an open interval. This is because the Riemann integral only considers the behavior of the function on a closed interval, while on an open interval, the function may have discontinuities or other behaviors that prevent it from being Riemann integrable.

4. What is the significance of a function being Riemann integrable?

A function being Riemann integrable is significant because it allows for the calculation of the area under the curve, which has many practical applications in mathematics and physics. In addition, the Riemann integral is a fundamental concept in calculus and is used to define more advanced integration techniques.

5. How does Riemann integrability differ from other types of integrability?

Riemann integrability differs from other types of integrability, such as Lebesgue integrability, in that it only considers the behavior of a function on a closed interval. Lebesgue integrability, on the other hand, considers the behavior of a function on the entire domain. Additionally, Riemann integrability requires the limit of the Riemann sums to exist, while Lebesgue integrability only requires the function to be measurable.

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