# Proof of a set is sigma finite

• manuel huant
In summary, the statement is saying that if a function f is integrable, then the set of points where the function is not equal to zero is a sigma-finite set. This means that the measure of this set can be expressed as a countable union of finite measure sets. The proof for this involves showing that for any n, the set of points where the function's absolute value is greater than 1/n has a finite measure.
manuel huant

## Homework Statement

if f is integrable, then the set N(f) = {x : f(x)≠ 0} is $\sigma$-finite

## Homework Equations

i am stucked in this proof , somebody help me please

## The 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:

## Homework Statement

if f is integrable, then the set N(f) = {x : f(x)≠ 0} is $\sigma$-finite

## Homework Equations

i am stucked in this proof , somebody help me please

## The 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?

What's the definition of N(f) being sigma finite? N(f) doesn't have to be finite. f(x)=1/x^2 is integrable on [1,infinity). [1,infinity) isn't finite.

Last edited:
sorry i didn't put it right , it seems N(f) $\sigma$-finite means the measure of N(f) is a countable union of finite measure sets
u(N(f))= $\underbrace{\cup}_{n}$u(N(fn)) which u(N(fn)) <$\infty$

manuel huant said:
sorry i didn't put it right , it seems N(f) $\sigma$-finite means the measure of N(f) is a countable union of finite measure sets
u(N(f))= $\underbrace{\cup}_{n}$u(N(fn)) which u(N(fn)) <$\infty$

Define a set A(f,n)={x: |f(x)|>1/n}. A(f,n) will have finite measure, right?

## 1. What is sigma finite?

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.

## 2. How is sigma finite different from finite?

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.

## 3. What is the importance of sigma finite sets in measure theory?

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.

## 4. How do you prove that a set is sigma finite?

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.

## 5. Are all measurable sets sigma finite?

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.

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