Home Work 1

Clearly has no upper bound

Hence,

is bounded as range of is First we need to show that $ε ε$

ε ε ε ε

ε ε ε ε

Hence for every $ε$ we can construct $ε$ $ε$ hence

if is constant than any will satisfy as so lets show the other way that is if there exist we need to show that is constant say

sum of non-positive terms is 0 iff all the terms are 0

this shows that has the same supremum and infimum hence is constant in hence is constant in

let (we are not claiming ) $ε$ for a small $ε$ $ε$ we are using the infimum and supriemum property to find

ε ε

as $ε$ is arbitrary

here is not integrable but is Riemann integrable

Fix then Let be sup in in and be inf in and be inf in clearly and

from the previous question we have seen if and hence

No, Let be any bounded function which is not Riemann integrable let be a constant and be a constant sup can be Dirichlet function This is a counter example proving the statement false

$ε ε ε ε$

ε ε ε ε ε ε ε ε

As a for every $ε$ $ε$ Hence is Riemann integrable

ε ε ε ε ε

For any as between any 2 real numbers there exist a irrational number and f(irrational number)=0 hence inf of any sub interval is 0 lets focus of $ε$ Let be let be

chose $ε$ once is fixed chose $ε$ so we get

ε ε ε

Hence $ε$

BMath. Notes

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Home Work 1

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