Chapter 1: Problem 13
Check that if \(f \geqslant 0\) and if \(\int_{Q} f=0\), then \(f=0\) a.e. Hint: $$ (x: f(x)>0)=\bigcup_{n \geqslant 1}(x: f(x)>1 / n) . $$
Short Answer
Expert verified
If \( f \geq 0 \) and \( \int_Q f = 0 \), then \( f = 0 \) almost everywhere.
Step by step solution
01
Understanding the Problem
We are tasked with showing that for a non-negative function \( f \), if the integral of \( f \) over the set \( Q \) equals zero, then \( f \) must be zero almost everywhere on \( Q \). The hint suggests examining the set where \( f(x) > 0 \) in terms of simpler sets \( (x: f(x) > 1/n) \).
02
Using the Hint
We start by considering the set where \( f(x) > 0 \). According to the hint, this set can be expressed as a countable union of sets where \( f(x) > 1/n \) for all \( n \geq 1 \). Mathematically, this is: \[ (x : f(x) > 0) = \bigcup_{n \geq 1} (x : f(x) > 1/n). \]
03
Integrating over the Sets
Since \( f \geq 0 \), if \( \int_Q f = 0 \), it implies that the measure of the set where \( f(x) > 0 \) must be zero. Now, consider the integral over the set \( (x: f(x) > 1/n) \). For each \( n \), \[ \int_{(x: f(x) > 1/n)} f(x) \, dx \leq \frac{1}{n} \cdot \text{measure}((x: f(x) > 1/n)). \] Since the full integral \( \int_Q f = 0 \), it means that the measure of each \( (x: f(x) > 1/n) \) must be zero for each \( n \).
04
Conclusion on the Measure
The set \( (x: f(x) > 0) \), being a countable union of measure zero sets \( (x: f(x) > 1/n) \), also has measure zero by the properties of measures. Hence, \( f(x) = 0 \) almost everywhere on \( Q \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Non-negative Functions
Non-negative functions play a critical role in measure theory. These are functions that satisfy the condition \( f(x) \geq 0 \) for all \( x \) within their domain. In simple terms, this means they never take negative values. When dealing with integrals involving non-negative functions, a natural property arises: the integral of a non-negative function is itself non-negative.
Consider a scenario where such a function is integrated over a set \( Q \) and the result is zero, symbolically presented as \( \int_Q f \, dx = 0 \). Even though the function \( f \) is not negative, this result suggests that \( f \) must be zero in almost all locations within the set \( Q \). If it were non-zero even at a single point with positive measure, the integral would not be zero.
Understanding non-negative functions is key to grasping further concepts in measure theory. They help simplify problems by limiting function behavior to only zeros and positive values.
Consider a scenario where such a function is integrated over a set \( Q \) and the result is zero, symbolically presented as \( \int_Q f \, dx = 0 \). Even though the function \( f \) is not negative, this result suggests that \( f \) must be zero in almost all locations within the set \( Q \). If it were non-zero even at a single point with positive measure, the integral would not be zero.
Understanding non-negative functions is key to grasping further concepts in measure theory. They help simplify problems by limiting function behavior to only zeros and positive values.
Measure Zero
In measure theory, a set is said to have measure zero if it is "small" in a certain rigorous sense. More formally, a set \( A \) has measure zero when, for every \( \epsilon > 0 \), there exists a series of intervals \((I_n)\) such that the set \( A \) is entirely contained within these intervals and the sum of their lengths is less than \( \epsilon \).
- Intuition: Think of measure zero sets as those that can be "covered" by very tiny intervals, such that their total size or length can be made arbitrarily small.
- Example: The set of all rational numbers within an interval like \([0,1]\) has measure zero, despite there being infinitely many rational numbers in the interval.
Countable Unions
The concept of countable unions is significant in analyzing set properties in measure theory. A countable union involves combining countably many sets to form a new set. Countable means the sets can be enumerated or listed in sequence (like natural numbers).
In the context of our exercise, the set where \( f(x) > 0 \) is expressed as a countable union of simpler sets \((x: f(x) > 1/n)\) across all positive integers \( n \). Each of these component sets may have measure zero, leading to an important conclusion:
In the context of our exercise, the set where \( f(x) > 0 \) is expressed as a countable union of simpler sets \((x: f(x) > 1/n)\) across all positive integers \( n \). Each of these component sets may have measure zero, leading to an important conclusion:
- Union Property: If each individual set in a countable collection has measure zero, their countable union also has measure zero.
- Realization: By expressing \( (x: f(x) > 0) \) as a countable union of \((x: f(x) > 1/n)\), we utilize their zero measure property.