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]$\rightarrow$R be a continuously differentable function
such that $\phi$'(t) $\geq$0 $\forall$t (i.e. $\phi$
is monotone increasing). Using only Reimann lemma, show that the composition $\phi$$\circ$f is Riemann integrable.

Homework Equations

Riemann lemma - f: [a, b] $\rightarrow$ is Riemann integrable iff for any $\epsilon$>0 $\exists$a 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$\rightarrow$n) $\sum$M_iΔx_i where Δx_i=[x_i-x_i-1]

The lower Riemann sum = i=(1$\rightarrow$n) $\sum$m_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

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

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


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

 

[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$\leq$l $\phi$(f(ci) - $\phi$(f(di)) l $\leq$2K?

 

[micromass]

Thus, 0$\leq$l $\phi$(f(ci) - $\phi$(f(di)) l $\leq$2K?

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)