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

[fabiancillo]

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}} $

 

[Jhoneine]

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

 

[soloenergy]

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