manuel huant said:Homework Statement
if f is integrable, then the set N(f) = {x : f(x)≠ 0} is [itex]\sigma[/itex]-finite
Homework Equations
i am stucked in this proof , somebody help me pleaseThe Attempt at a Solution
if f is simple the it seems the set is finite since otherwise the the integral won't exist but how can it be extended to f is integrable?
manuel huant said:sorry i didn't put it right , it seems N(f) [itex]\sigma[/itex]-finite means the measure of N(f) is a countable union of finite measure sets
u(N(f))= [itex]\underbrace{\cup}_{n}[/itex]u(N(fn)) which u(N(fn)) <[itex]\infty[/itex]
Sigma finite is a property of a set in measure theory, which means that the set can be written as a countable union of measurable subsets with finite measure. In other words, the set can be "cut up" into smaller parts that have a finite size.
A set that is finite has a finite number of elements, while a set that is sigma finite can have an infinite number of elements, but they must be able to be grouped into a countable number of subsets with finite measure. Sigma finite sets can be considered as "infinite but manageable" compared to truly infinite sets.
Sigma finite sets are important in measure theory because they allow for the creation of measures on larger, more complicated sets. This is because sigma finite sets can be "approximated" by finite sets, making them easier to work with mathematically.
To prove that a set is sigma finite, you must show that it can be written as a countable union of measurable subsets with finite measure. This can be done by constructing a sequence of sets with increasing measure that ultimately "covers" the original set.
No, not all measurable sets are sigma finite. For example, sets with infinite measure (such as the real numbers) are not sigma finite. However, many important sets in measure theory, such as intervals and rectangles, are sigma finite.