Lebesgue measurable sets

  • Thread starter To0ta
  • Start date
  • #1
6
0
let [tex]\mu^{}*[/tex] , v[tex]^{}*[/tex] outer measura on X . Show that max{[tex]\mu^{}*[/tex] , v[tex]^{}*[/tex]} is an outer measure on X ?
 

Answers and Replies

  • #2
237
0
The first (possibly only) thing to try would be to look at the properties that define an outer measure, and check whether max(u,v) satisfies them. Did you try that yet?
 
  • #3
6
0
The first (possibly only) thing to try would be to look at the properties that define an outer measure, and check whether max(u,v) satisfies them. Did you try that yet?

Yes I have tried a lot
 
  • #4
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,872
416
Yes I have tried a lot
And what was the result?
 
  • #5
104
2
FAIL.

From wikipedia: http://en.wikipedia.org/wiki/Outer_measure#Formal_definitions

Defining properties of an outer measure:

* The empty set has measure 0.
* Monotonicity: If A is a subset of B, then the measure of A is at most the measure of B.
* Countable Subadditivity: The measure of a countable union of sets is at most the sum of the measures of each of the sets in the union.

If u* and v* are outer measures, then max{u*,v*} is outer measure if and only if it satisfies the above three properties.

In other words:

* max{u*(empty set), v*(empty set)} = 0.

* If A is a subset of B, then max{u*A,v*A} is less than or equal to max{u*B,v*B}.

*If A1, A2, A3, ... are sets, and A is their union, then max{u*A,v*A} is less than or equal to the sum over i = 1,2,3,... of max{u*Ai,v*Ai}.

The first two conditions are really, really straightforward. The third follows from the fact that the maximum of two sums (say, for example, of max{sum of x_i, sum of y_i}) is at most the sum of the maximums (i.e. the sum of max{x_i,y_i}).
 
  • #6
6
0
FAIL.

From wikipedia: http://en.wikipedia.org/wiki/Outer_measure#Formal_definitions

Defining properties of an outer measure:

* The empty set has measure 0.
* Monotonicity: If A is a subset of B, then the measure of A is at most the measure of B.
* Countable Subadditivity: The measure of a countable union of sets is at most the sum of the measures of each of the sets in the union.

If u* and v* are outer measures, then max{u*,v*} is outer measure if and only if it satisfies the above three properties.

In other words:

* max{u*(empty set), v*(empty set)} = 0.

* If A is a subset of B, then max{u*A,v*A} is less than or equal to max{u*B,v*B}.

*If A1, A2, A3, ... are sets, and A is their union, then max{u*A,v*A} is less than or equal to the sum over i = 1,2,3,... of max{u*Ai,v*Ai}.

The first two conditions are really, really straightforward. The third follows from the fact that the maximum of two sums (say, for example, of max{sum of x_i, sum of y_i}) is at most the sum of the maximums (i.e. the sum of max{x_i,y_i}).

thanks

Can you resolved by using with another idea





Max{a,b}=a+b+|a-b| / 2
 

Related Threads on Lebesgue measurable sets

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
7
Views
5K
  • Last Post
Replies
5
Views
8K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
8
Views
2K
Replies
2
Views
2K
Top