Associated Topics || Dr. Math Home || Search Dr. Math

### Riemann Integrals, etc.

```
Date: 24 Apr 1995 23:12:25 -0400
From: Anonymous
Subject: first of three problems

Hi ..this has something to do with the Riemann Integral..please
give me a hint on how to start this!

Let A be a subset of the Reals, Prove that:
i) If A is finite, then the outer measure is zero
ii) If A = [a,b], then the outer measure = (b - a)

Next problem:

Let f and g be elements of the Reals, and let I be a subset of the
Reals.

{ x is an element of I such that f(x) does not equal g(x)} is finite

Then the integral of f, from a to b equals the integral of g
from a to b
```

```
Date: 28 Apr 1995 15:23:31 -0400
From: Dr. Ken
Subject: Re: first of three problems

Hello there!

>Let A be a subset of the Reals, Prove that:
>i) If A is finite, then the outer measure is zero

The key thing in showing that the outer measure for some set is zero is to
show that the measure is less than every possible positive number.  Let e
(epsilon) be a small positive number.  See whether you can show that there
is some covering of A that has measure less than e.  Hint: if there are n
points in the set, how big can the covering around each point be so that the
total covering is less than e?

>ii) If A = [a,b], then the outer measure = (b - a)

This is a similar problem, and I think it will be easier to do once you've
done the first part.  If not, write back and we'll try to help you out.

>let f and g be elements of the Reals ,and let I be a subset of the
>Reals.
>
> { x is an element of I such that f(x) does not equal g(x)} is finite
>
>Then    the integral of f, from a to b equals the integral of g
>          from a to b

You (hopefully) showed that a finite set of points in the real number has
outer measure zero.  There's a theorem that says that if two functions only
disagree on a set of measure zero in an interval, then their integrals on
that interval are equal.  To prove that, you could break up the integrals
into lots of smaller integrals, where some of the integrals have only points
where f and g agree, and some of the integrals are centered around points
where f and g disagree.  Then show how you can make the total area found
with the second kind of integrals really small.

-Ken "Dr." Math
```

```
Date: 29 Apr 1995 11:18:52 -0400
From: Anonymous
Subject: Re: first of three problems

Hi
Thanks for the hints, I'll do that and get back to you..
Here's the last problem

Suppose (f sub n)n converges uniformly to f.
Then, If each function f sub n is bounded, so is f
Thanks
Mike
```

```
Date: 30 Apr 1995 12:38:36 -0400
From: Dr. Sydney
Subject: Re: first of three problems

Dear Mike,

Hi!  Good question.  I'll try to give you a few ideas for directions in
which you might go, and maybe you can take it from there...

Write down the explicit definitions of uniform convergence of f(sub n) to f
and bounded f...

Now maybe you can work with a triangle inequality in which f (sub n) and f
are involved.  |f(x)| = |f(x) - f(sub n)(x) + f(subn)(x)|, right?  Now make
this into a triangle inequality, and then apply the definition of uniform
convergence for a specific epsilon, say epsilon = 1/2.

If these hints are too vague or are confusing, or if you want to check your
answer just let us know!  Good luck!

--Sydney, "Dr. Math"
```
Associated Topics:
College Analysis

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search