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

[IntroAnalysis]

Homework Statement

Let f:[a, b]\rightarrow[m, M] be a Riemann integrable function and let
\phi:[m, M]\rightarrowR be a continuously differentable function
such that \phi'(t) \geq0 \forallt (i.e. \phi
is monotone increasing). Using only Reimann lemma, show that the composition \phi\circf is Riemann integrable.

Homework Equations

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

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

Also, since the domain of \phi 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 \phi is also bounded. Thus, \phi((f(a)) and \phi(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)< \epsilon

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

I think I have most of the major pieces, can someone suggest how to put it together?
Thank you.

 

[micromass]

Why don't you start by writing out the definition of

U(P,f)-L(P,f)

 

[IntroAnalysis]

The upper Riemann sum = (i=1\rightarrown) \sumM_iΔx_i where Δx_i=[x_i-x_i-1]

The lower Riemann sum = i=(1\rightarrown) \summ_iΔx_i where Δx_i=[x_i-x_i-1]

M_i=sup[f(x_i): x\in[xi, xi-1]
m_i=inf[f(x_i): x\in[xi, xi-1]

 

[micromass]

OK, let c_i (resp. d_i) be the element of [x_i,x_{i+1}], where f reaches his maximum (resp. minimum).

We know that

\sum_{i=1}^n (f(c_i)-f(d_i)) \Delta x_i&lt;\varepsilon

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

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


\sum_{i=1}^n (\varphi (f(c_i))-\varphi(f(d_i))) \Delta x_i&lt;\varepsilon

 

[IntroAnalysis]

Can't we just say that since \phi is monotone increasing, that we know that \phi(f(ci)) (resp. \phi(f(di))) is where \phi reaches its maximum (resp. minimum)?

Thus, 0\leql \phi(f(ci) - \phi(f(di)) l \leq2K?

 

[micromass]

Thus, 0\leql \phi(f(ci) - \phi(f(di)) l \leq2K?

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

But, basically, we have something of the form

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

and you must associate this with

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

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