Proving Measure Space Properties of $(X,\bar{\mathcal{B}} ,\bar{\mu})$

In summary: B}}$ is a subset of $\bar{\mu}(X)$. To do this, we first note that $\mathcal{B}$ is a subset of $\bar{\mu}(X)$ since $\mathcal{B} \subseteq \bar{\mu}(X)$ for all $X \in \mathbb{R}^n$. Next, we observe that $\bar{\mathcal{B}}$ is a subset of $\bar{\mu}(X)$ since $\bar{\mathcal{B}} \subseteq \bar{\mu}(X)$ if and only if $\mu(\mathcal{B}) =
  • #1
fabiancillo
27
1
Hello, I have problems with this exercise

Let $(X,\mathcal{B} , \mu)$ a measurement space, consider

$\bar{\mathcal{B}} = \{ A \subseteq{X} \; : \; A\cap{B} \in \mathcal{B}$ for all that satisfies $\mu(B) < \infty \}$, and

for $A \in \bar{\mathcal{B}}$ define

$\bar{\mu}(A) = \left \{ \begin{matrix} \mu (A) & \mbox{if }A \in \mathcal{B}
\\ +\infty & \mbox{if }A \not\in\mathcal{B}\end{matrix}\right. $

Prove that $(X,\bar{\mathcal{B}} ,\bar{\mu})$ X is a measure space and $\mathcal{B} \subseteq{} \bar{\mathcal{B}} $
 
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  • #2
Solution: First, it is clear that $\mathcal{B} \subseteq{} \bar{\mathcal{B}}$ since $\mathcal{B}$ is a subset of $\bar{\mathcal{B}}$. Next, we need to prove that $(X,\bar{\mathcal{B}} ,\bar{\mu})$ is a measure space. To do this, we must show that $\bar{\mu}$ is a measure on $\bar{\mathcal{B}}$. We first note that $\bar{\mu}$ is non-negative, since $\bar{\mu}(A) \geq 0$ for all $A \in \bar{\mathcal{B}}$. Next, we need to show that $\bar{\mu}$ is countably additive. To show this, let $A_1, A_2, \ldots$ be disjoint sets in $\bar{\mathcal{B}}$. We note that since $\mu$ is countably additive, $\mu(\bigcup\limits_{i=1}^{\infty}A_i) = \sum\limits_{i=1}^{\infty}\mu(A_i)$. Since each $A_i \in \bar{\mathcal{B}}$, we have that $A_i \cap B \in \mathcal{B}$ for all $B$ with $\mu(B) < \infty$. Thus, $\mu(\bigcup\limits_{i=1}^{\infty}A_i \cap B) = \sum\limits_{i=1}^{\infty}\mu(A_i \cap B)$ for all such $B$. It follows that $\bar{\mu}(\bigcup\limits_{i=1}^{\infty}A_i) = \mu(\bigcup\limits_{i=1}^{\infty}A_i) = \sum\limits_{i=1}^{\infty}\mu(A_i) = \sum\limits_{i=1}^{\infty}\bar{\mu}(A_i)$, which shows that $\bar{\mu}$ is indeed countably additive. Finally, we need
 
  • #3
Hi there,

I'm sorry to hear that you're having trouble with this exercise. Let me try to help you out.

First, to prove that $(X,\bar{\mathcal{B}},\bar{\mu})$ is a measure space, we need to show that it satisfies the three properties of a measure: non-negativity, countable additivity, and null empty set.

1. Non-negativity: Since $\mu$ is a measure, we know that it is non-negative, meaning that $\mu(A) \geq 0$ for all $A \in \mathcal{B}$. This property also holds for $\bar{\mu}$, since $\bar{\mu}(A) = \mu(A)$ for $A \in \mathcal{B}$ and $\bar{\mu}(A) = +\infty$ for $A \not\in\mathcal{B}$.

2. Countable additivity: Let $A_1, A_2, \dots$ be a countable collection of pairwise disjoint sets in $\bar{\mathcal{B}}$. We want to show that $\bar{\mu}(\bigcup_{i=1}^{\infty} A_i) = \sum_{i=1}^{\infty} \bar{\mu}(A_i)$.

First, note that for any $B \in \mathcal{B}$ with $\mu(B) < \infty$, we have $A_i \cap B \in \mathcal{B}$ since $A_i \in \bar{\mathcal{B}}$ and $B \in \mathcal{B}$. Therefore, by the countable additivity property of $\mu$, we have $\mu(\bigcup_{i=1}^{\infty} (A_i \cap B)) = \sum_{i=1}^{\infty} \mu(A_i \cap B)$.

Now, for any $A_i \not\in \mathcal{B}$, we have $\bar{\mu}(A_i) = +\infty$. So, $\mu(A_i \cap B) \leq \mu(B) < \infty$ for all $i$, and therefore $\sum_{i=1}^{\infty} \mu(A_i \cap B) < \infty$. This means that
 

Related to Proving Measure Space Properties of $(X,\bar{\mathcal{B}} ,\bar{\mu})$

1. What is a measure space?

A measure space is a mathematical concept used in probability theory and measure theory. It consists of a set X, a sigma-algebra &bar;B of subsets of X, and a measure &bar;μ that assigns a non-negative number to each element of &bar;B.

2. How do you prove that a set is a sigma-algebra?

To prove that a set is a sigma-algebra, you must show that it satisfies three properties: closure under complementation, closure under countable unions, and containing the empty set. This can be done by using logical reasoning and the definitions of these properties.

3. What is the difference between a measure and a sigma-algebra?

A measure is a function that assigns a non-negative number to a set, while a sigma-algebra is a collection of sets that satisfies certain properties. In a measure space, the sigma-algebra determines which sets can be assigned a measure, and the measure determines the size or weight of those sets.

4. How do you prove that a measure is finite or infinite?

To prove that a measure is finite, you must show that the measure of the entire space X is a finite number. To prove that a measure is infinite, you must show that the measure of at least one set in the sigma-algebra is infinite.

5. What is the importance of proving measure space properties?

Proving measure space properties is important because it allows us to understand the behavior of measures and sets in a mathematical context. It also allows us to apply these properties to various fields, such as probability theory and analysis, to make more accurate and meaningful conclusions.

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