Riemann integral and null sets

• Jan 28th 2010, 04:47 PM
iknowone
Riemann integral and null sets
Let $f$ be a nonnegative Riemann integrable function on [a,b].

Prove that $int_{a}^{b}f(x) dx = 0$

if and only if

$E = {x \in [a,b] | f(x) > 0}$ is a null set.
___________________

<= If E is a null set I want to be able to define a partition
that allows me to control the contribution to the integral on those
intervals which contain points in E by arguing that these intervals have
arbitrarily small length and f(x) is bounded. Then I can control the
contribution to the integral on those intervals which do not contain
points in E by observing that the function is equal to 0 at these points.
But given that E is a null set I get a cover of E with arbitrarily small
length. But this cover is the countable union of intervals. I need to refine
this to a finite subcover if I hope to construct a partition as described
above. E needn't be compact as far as I can tell thus a finite subcover is
not guaranteed.

=> No idea.
• Jan 28th 2010, 04:59 PM
Defunkt
Nevermind, was unfamiliar with the 'null set' terminology.
• Jan 28th 2010, 05:01 PM
TheEmptySet
Quote:

Originally Posted by iknowone
Let $f$ be a nonnegative Riemann integrable function on [a,b].

Prove that $int_{a}^{b}f(x) dx = 0$

if and only if

$E = {x \in [a,b] | f(x) > 0}$ is a null set.
___________________

<= If E is a null set I want to be able to define a partition
that allows me to control the contribution to the integral on those
intervals which contain points in E by arguing that these intervals have
arbitrarily small length and f(x) is bounded. Then I can control the
contribution to the integral on those intervals which do not contain
points in E by observing that the function is equal to 0 at these points.
But given that E is a null set I get a cover of E with arbitrarily small
length. But this cover is the countable union of intervals. I need to refine
this to a finite subcover if I hope to construct a partition as described
above. E needn't be compact as far as I can tell thus a finite subcover is
not guaranteed.

=> No idea.

Correct me if I'm wrong but I don't think this is true

Set $E=\mathbb{Q} \cap [a,b]$

Since the rational numbers are countable this mean we can we can index them with the natural numbers and consider the denumeration

$q_1,q_2,q_3,....$
and let $\epsilon > 0$ then

$E \subset Q$ but

$Q \subset \bigcup_{n=1}^{\infty}(q_n-\frac{\epsilon}{2^{n+1}},q_n+\frac{\epsilon}{2^{n+ 1}})$

This last set have measure zero i.e it is a null set and contains E.
• Jan 28th 2010, 05:09 PM
Jose27
Quote:

Originally Posted by iknowone
Let $f$ be a nonnegative Riemann integrable function on [a,b].

Prove that $int_{a}^{b}f(x) dx = 0$

if and only if

$E = {x \in [a,b] | f(x) > 0}$ is a null set.
___________________

<= If E is a null set I want to be able to define a partition
that allows me to control the contribution to the integral on those
intervals which contain points in E by arguing that these intervals have
arbitrarily small length and f(x) is bounded. Then I can control the
contribution to the integral on those intervals which do not contain
points in E by observing that the function is equal to 0 at these points.
But given that E is a null set I get a cover of E with arbitrarily small
length. But this cover is the countable union of intervals. I need to refine
this to a finite subcover if I hope to construct a partition as described
above. E needn't be compact as far as I can tell thus a finite subcover is
not guaranteed.

=> No idea.

$\Rightarrow$ ) If $\mu (E) >0$ then $0< \int_{E} f \leq \int_{a}^{b} f$ which is a contradiction.

$\Leftarrow$ ) If $E$ is a null set then $\int_{a}^{b} f=\int_{a}^{b} 0=0$ (because the integral doesn't see null sets)

One question though: Have you seen the Lebesgue integral? because otherwise my 'proof' might not be so adequate.
• Jan 30th 2010, 03:45 PM
iknowone
No, I want to show this result using only the definition of the Riemann integral as the inf of the set upper sums over all partitions of [a,b].