1. ## HELP!! thankx

3 questions:
1) let f: (a,b) -> (c,d) be continuos surjective monotonic function. prove f is invertible and its inverse is continuous

2) f: [0,1] -> R
f(x) = 0 if x is irrational;
f(x) = 1/n if x is rational with x = m/n and gcd(m,n) = 1
prove that f is riemann integrable and its integral from 0 to 1 is 0

3) let f(x) < g(x) < h(x) for all x in [a,b]
suppose f and h are riemann intagrable and integral of f from 0 to 1 equals integral of h from 0 to 1
prove g is riemann intagrable and integral of g from 0 to 1 equals integral of f from 0 to 1

2. Originally Posted by pc31
3 questions:

2) f: [0,1] -> R
f(x) = 0 if x is irrational;
f(x) = 1/n if x is rational with x = m/n and gcd(m,n) = 1
prove that f is riemann integrable and its integral from 0 to 1 is 0
Before we get started lets note a few things. $f$ is bounded above on $[0,1]$ by 1. There is an irrational and rational point in every interval of postive length.

Let $\epsilon > 0$ Pick an $N \in \mathbb{N}$ such that $\frac{1}{N} < \frac{\epsilon }{4}$.

Now lets construct a partion of [0,1]. let n > N

$A=\left[ 0,\frac{1}{n}\right]$

$B=\bigcup_{i=2}^{n}\left[ \frac{1}{i}-\frac{\epsilon}{2^{i+1}},\frac{1}{i}+\frac{\epsilo n}{2^{i+1}}\right]$

let $C=\left[ 1-\frac{1}{n},1\right]$

Now set $P=[A \cup B \cup C]$

$|U(f,P)-L(f,P)|$
$< \left|1 \cdot \left[ \frac{1}{n}-0 \right]+\sum_{i=2}^{n} 1 \cdot \left( \frac{1}{n} + \frac{ \epsilon}{2^{i+1}} - \frac{1}{n} + \frac{\epsilon}{2^{i+1}} \right) +1 \cdot \left( 1 -1+\frac{1}{n} \right) \right|$

$=\left| \frac{2}{n}+\epsilon \sum_{i=2}^{n}\left( \frac{1}{2^i}\right) \right|<2\cdot \frac{\epsilon}{4}+\frac{\epsilon}{2}=\epsilon$

So f is integrable on [0,1]. Since it is integrable the value of the integral is equal to the supreemum of the lower integrals. Since it is always zero the value of the integral is 0.

Good luck.

3. For 3) use the squeeze theorem.

4. Originally Posted by pc31
3 questions:
1) let f: (a,b) -> (c,d) be continuos surjective monotonic function. prove f is invertible and its inverse is continuous
First note, that if $f$ is a constant function, we have the result right off the bat, so lets assume it is not constant, that is, it is increasing or decreasing. Now, a function is invertible if it is both 1-1 and onto. we are given that $f$ is onto, thus, to prove it is invertible, it remains to show it is 1-1. we will do this using the definition: A function $f$ is 1-1 if $x_1 \ne x_2 \implies f(x_1) \ne f(x_2)$ for $x_1,x_2 \in \text{dom}(f)$.

Assume $x_1 \ne x_2$. Then, we may assume, without loss of generality, that $x_2 > x_1$, and that $f$ is an increasing function (if this is not the case, we can easily assume otherwise and the proof remains valid). Since $f$ is montone, $x_2 > x_1$ implies that $f(x_2) > f(x_1)$. But that means $f(x_1) \ne f(x_2)$, as desired.

Now to prove continuity of the inverse function (i think this works ).

Let $\{ x_n \}$ be a convergent sequence in $\text{dom}(f)$. and assume $\lim_{n \to \infty} f(x_n) = L$. this is possible by the continuity of $f$. by the Heine definition of continuity, we need to show that $\lim_{n \to \infty} f^{-1}(f(x_n)) = f^{-1}(L)$. But that is, $\lim_{n \to \infty} x_n = f^{-1}(L)$. this follows by our original assumption and the continuity of $f$.

(Note that the domain of the inverse function is the codomain of the function. thus, for $x$ in the domain of $f$, it makes sense to write $f(x)$ to represent an element in the domain of $f^{-1}(x)$)