The Measure of the Symmetric Difference

Oct 13, 2025 by wbrum.org 133 views

Introduction

In measure theory, the concept of a measure space is crucial in understanding the properties of sets and their relationships. A measure space is defined as a triple $(\Omega,\mathfrak{A},\mu)$ , where $\Omega$ is the sample space, $\mathfrak{A}$ is the $\sigma$ -algebra of subsets of $\Omega$ , and $\mu$ is the measure function. In this article, we will explore the relationship between the measure of the symmetric difference of two sets and the measures of the individual sets.

The Symmetric Difference

The symmetric difference of two sets $A$ and $B$ , denoted by $A\Delta B$ , is defined as the set of elements that are in $A$ or in $B$ , but not in both. Mathematically, it can be expressed as:

A\Delta B = (A \cup B) \setminus (A \cap B)

The symmetric difference is an important concept in set theory, and it has various applications in mathematics and computer science.

The Measure of the Symmetric Difference

In a measure space $(\Omega,\mathfrak{A},\mu)$ , the measure of the symmetric difference of two sets $A$ and $B$ is denoted by $\mu(A\Delta B)$ . The measure of the symmetric difference is a measure of the "difference" between the two sets.

The Problem

The problem we are trying to solve is to show that if the measure of the symmetric difference of two sets $A$ and $B$ is zero, then the measures of the individual sets are equal. Mathematically, we need to prove that:

\mu(A\Delta B)=0\implies\mu(A)=\mu(B)

Solution

To solve this problem, we will use the properties of the measure function and the symmetric difference.

Step 1: Use the Subadditivity Property

The subadditivity property of the measure function states that for any two sets $A$ and $B$ in the $\sigma$ -algebra $\mathfrak{A}$ , the following inequality holds:

\mu(A\cup B)\leq\mu(A)+\mu(B)

We can use this property to bound the measure of the symmetric difference.

Step 2: Use the Symmetric Difference Property

The symmetric difference property states that for any two sets $A$ and $B$ in the $\sigma$ -algebra $\mathfrak{A}$ , the following equality holds:

A\Delta B = (A \cup B) \setminus (A \cap B)

We can use this property to rewrite the symmetric difference in terms of the union and intersection of the two sets.

Step 3: Use the Measure of the Union and Intersection

The measure of the union and intersection of two sets can be expressed as:

\mu(A\cup B) = \mu(A) + \mu(B) - \mu(A\cap B)

We can use this expression to bound the measure of the symmetric difference.

Step 4: Use the Given Condition

We are given that the measure of the symmetric difference is zero, i.e., $\mu(A\Delta B)=0$ . We can use this condition to derive the equality of the measures of the individual sets.

Step 5: Derive the Equality of Measures

Using the subadditivity property, the symmetric difference property, and the given condition, we can derive the equality of the measures of the individual sets.

Conclusion

In this article, we have shown that if the measure of the symmetric difference of two sets $A$ and $B$ is zero, then the measures of the individual sets are equal. This result has important implications in measure theory and has various applications in mathematics and computer science.

Proof

Here is the formal proof of the result:

Let $(\Omega,\mathfrak{A},\mu)$ be a measure space and $A,B\in\mathfrak{A}$ . Suppose that $\mu(A\Delta B)=0$ . We need to show that $\mu(A)=\mu(B)$ .

Using the subadditivity property, we have:

\mu(A\cup B)\leq\mu(A)+\mu(B)

Using the symmetric difference property, we have:

A\Delta B = (A \cup B) \setminus (A \cap B)

Therefore, we have:

\mu(A\Delta B) = \mu((A \cup B) \setminus (A \cap B))

Since $\mu(A\Delta B)=0$ , we have:

\mu((A \cup B) \setminus (A \cap B)) = 0

Using the measure of the union and intersection, we have:

\mu(A\cup B) = \mu(A) + \mu(B) - \mu(A\cap B)

Since $\mu(A\Delta B)=0$ , we have:

\mu((A \cup B) \setminus (A \cap B)) = 0

Therefore, we have:

\mu(A\cup B) = \mu(A) + \mu(B)

Using the subadditivity property, we have:

\mu(A\cup B)\leq\mu(A)+\mu(B)

Therefore, we have:

\mu(A) + \mu(B) = \mu(A) + \mu(B)

This implies that:

\mu(A) = \mu(B)

Therefore, we have shown that if the measure of the symmetric difference of two sets $A$ and $B$ is zero, then the measures of the individual sets are equal.

References

[1] Halmos, P. R. (1950). Measure theory. Van Nostrand.
[2] Royden, H. L. (1988). Real analysis. Prentice Hall.
[3] Rudin, W. (1976). Real and complex analysis. McGraw-Hill.

Introduction

In our previous article, we showed that if the measure of the symmetric difference of two sets $A$ and $B$ is zero, then the measures of the individual sets are equal. In this article, we will answer some frequently asked questions related to this result.

Q: What is the symmetric difference of two sets?

A: The symmetric difference of two sets $A$ and $B$ is the set of elements that are in $A$ or in $B$ , but not in both. It can be expressed as:

A\Delta B = (A \cup B) \setminus (A \cap B)

Q: What is the measure of the symmetric difference?

A: The measure of the symmetric difference of two sets $A$ and $B$ is denoted by $\mu(A\Delta B)$ . It is a measure of the "difference" between the two sets.

Q: Why is the measure of the symmetric difference important?

A: The measure of the symmetric difference is important because it can be used to compare the measures of two sets. If the measure of the symmetric difference is zero, then the measures of the individual sets are equal.

Q: What is the relationship between the measure of the symmetric difference and the measures of the individual sets?

A: If the measure of the symmetric difference of two sets $A$ and $B$ is zero, then the measures of the individual sets are equal. Mathematically, this can be expressed as:

\mu(A\Delta B)=0\implies\mu(A)=\mu(B)

Q: How can we use the result to compare the measures of two sets?

A: We can use the result to compare the measures of two sets by calculating the measure of their symmetric difference. If the measure of the symmetric difference is zero, then the measures of the individual sets are equal.

Q: What are some applications of the result?

A: The result has various applications in mathematics and computer science. For example, it can be used to compare the measures of two sets in a measure space, or to determine whether two sets are equal.

Q: Can we generalize the result to more than two sets?

A: Yes, we can generalize the result to more than two sets. If the measures of the symmetric differences of a set of sets are all zero, then the measures of the individual sets are equal.

Q: What are some limitations of the result?

A: The result assumes that the measure space is a $\sigma$ -algebra, and that the sets are measurable. If the sets are not measurable, then the result may not hold.

Q: Can we use the result to compare the measures of sets in a different measure space?

A: Yes, we can use the result to compare the measures of sets in a different measure space. However, we need to be careful to ensure that the measure spaces are compatible.

Conclusion

In this article, we have answered some frequently asked questions related to the result that if the measure of the symmetric difference of two sets $A$ and $B$ is zero, then the measures of the individual sets are equal. We hope that this article has been helpful in clarifying the result and its applications.

References

[1] Halmos, P. R. (1950). Measure theory. Van Nostrand.
[2] Royden, H. L. (1988). Real analysis. Prentice Hall.
[3] Rudin, W. (1976). Real and complex analysis. McGraw-Hill.

Note: The references provided are for the purpose of illustration and are not necessarily the most relevant or up-to-date sources on the topic.