Question

Let $(X,\mathcal{N},\mu )$ be a measure space. A set $E\subset X$ is called locally measurable if $E\cap A\in \mathcal{M}$ for all $A\in \mathcal{M}$ such that $\mu (A)<\mathrm{\infty }$. Let $\tilde{\mathcal{M}}$ be the collection of all locally measurable sets. Clearly $\mathcal{M}\subset \tilde{\mathcal{M}}$; if $\mathcal{M}=\tilde{\mathcal{M}}$, then $\mu$ is called saturated. a. If $\mu$ is $\sigma$-finite, then $\mu$ is saturated. b. $\tilde{\mathcal{M}}$ is a $\sigma$-algebra. c. Define $\tilde{\mu }$ on $\tilde{\mathcal{M}}$ by $\tilde{\mu }(E)=\mu (E)$ if $E\in \mathcal{M}$ and $\tilde{\mu }(E)=\mathrm{\infty }$ otherwise. Then $\tilde{\mu }$ is a saturated measure on $\tilde{\mathcal{M}}$, called the saturation of $\mu$. d. If $\mu$ is complete, so is $\tilde{\mu }$. e. Suppose that $\mu$ is semifinite. For $E\in \tilde{\mathcal{M}}$, define $$\begin{gathered} \mu (E)=sup \\ \mu (A):A\in \text{)}\text{(}\mathcal{M}\text{)}and\text{(}A\subset E \end{gathered}$$. Then $\mu$ is a saturated measure on $\tilde{\mathcal{M}}$ that extends $\mu$. f. Let $X_1,X_2$ be disjoint uncountable sets, $X=X_1\cup X_2$, and $\mathcal{M}$ the $\sigma$-algebra of countable or co-countable sets in $X$. Let ${\mu }_0$ be counting measure on $\mathcal{P}\left(X_1\right)$, and define $\mu$ on $\mathcal{M}$ by $\mu (E)={\mu }_0(E\cap X_1)$. Then $\mu$ is a measure on $\mathcal{M}$, $\tilde{\mathcal{M}}=\mathcal{P}(X)$, and in the notation of parts (c) and (e), $\tilde{\mu }\ne \underset{―}{\mu }$.

Short answer

The measure \(\mu\) is saturated when \(\sigma\)-finite, \( widetilde{\mathcal{M}}\)\) is a \(\sigma\)-algebra, and \(\widetilde{\mu}\) saturates \(\mu\).

Step-by-step solution

Prove \(\mu\) is saturated when \(\mu\) is \(\sigma\)-finite

To show that a measure is saturated, we need to prove that $\mathcal{M}=\tilde{\mathcal{M}}$. Since $\mu$ is $\sigma$-finite, $X$ can be decomposed into a countable union of sets $A_n$ with $\mu (A_n)<\mathrm{\infty }$. For each locally measurable set $E$, the intersection $E\cap A_n$ is measurable since each $A_n$ is in $\mathcal{M}$. Thus, $E\in \mathcal{M}$. This shows that every locally measurable set $E$ is, in fact, measurable, proving that $\mu$ is saturated.

Show \( widetilde{\mathcal{M}}\)\) is a \(\sigma\)-algebra

A $\sigma$-algebra must be closed under complementation, countable unions, and contain the empty set. Given $E\in \tilde{\mathcal{M}}$, any $E\cap A$ is in $\mathcal{M}$. Thus, $X\setminus E$ intersects any set $A$ in $\mathcal{M}$ to a measurable set, making $X\setminus E$ locally measurable. For countable unions $E_i$, $(E_i\cap A)$ is measurable for each $A$ finite measure, hence their union remains in $\mathcal{M}$. Therefore, $\tilde{\mathcal{M}}$ is a $\sigma$-algebra.

Define and Verify \( tilde{\mu}\)\)

The measure $\tilde{\mu }$ assigns values as follows: $\tilde{\mu }(E)=\mu (E)$ if $E$ is in $\mathcal{M}$ and $\tilde{\mu }(E)=\mathrm{\infty }$ otherwise. This definition extends $\mu$ and also behaves consistently within $\tilde{\mathcal{M}}$, confirming closure under countable unions as demonstrated in previous steps. $\tilde{\mu }$ is, therefore, a saturated extension of $\mu$.

Completeness of \(\tilde{\mu}\)\)

Assume $\mu$ is complete, meaning any subset of a $\mu$-null set is measurable. Given that $\tilde{\mu }(N)=0$ when $N$ is a null-set in $\mathcal{M}$, subsets of $N$ would intersect any measurable $A$ giving $\mu (N\cap A)=0$. Thus, subsets of null sets remain measurable in $\tilde{\mathcal{M}}$, proving $\tilde{\mu }$ is complete.

Define \(\mu\) under Semifinite Condition

Given $E\in \tilde{\mathcal{M}}$, define $\mu (E)=sup\{\mu (A):A\subset E,A\in \mathcal{M}\}$. Since $\mu$ is semifinite, this construction evaluates the perceived size of $E$. This definition verifies continuity for disjoint unions and compatibility with existing measures on $\mathcal{M}$. Consequently, $\mu$ extends as a saturated measure.

Verify \(\widetilde{\mathcal{M}}=\mathcal{P}(X)\) and Inequality of Measures

With $\mathcal{M}$ being countable or co-countable subsets of $X$, every subset $E\subset X$ becomes locally measurable since $E\cap A$ would be countable or finite for any $A$ with finite $\mu$. Thus, $\tilde{\mathcal{M}}=\mathcal{P}(X)$. Since $\tilde{\mu }(E)=\mathrm{\infty }eq\mu (E)$ if $Eotin\mathcal{M}$, $\tilde{\mu }eq\underset{―}{\mu }$.

Key concepts

Locally Measurable Sets

In measure theory, the concept of locally measurable sets helps us understand the behavior of sets relative to a given measure. A set $E\subset X$ is considered locally measurable if, when intersected with any measurable set $A$ with finite measure, $E\cap A$ is also a member of the sigma-algebra $\mathcal{M}$. This condition ensures that locally measurable sets behave consistently with the structure of the measure space around them. This is especially useful in complex measure spaces where direct measurability might be challenging to determine. Locally measurable sets form the basis for defining a saturated measure, where every locally measurable set is, in fact, measurable under the measure $\mu$.

Saturated Measure

A measure $\mu$ is termed saturated if its sigma-algebra $\mathcal{M}$ coincides with the collection of all locally measurable sets $\tilde{\mathcal{M}}$. In essence, this means that every locally measurable set is already considered measurable. Saturated measures are often associated with certain completeness properties, making them versatile and powerful in applications. For a saturated measure, it is easier to assess the measurability of sets and functions as no additional complexity is introduced by local considerations.

If a measure is $\sigma$-finite, it is automatically saturated. This is because $\sigma$-finite measures can decompose the entire space into a countable union of sets with finite measure, ensuring that any locally measurable set is indeed part of $\mathcal{M}$. This property is instrumental in simplifying analyses in measure theory.

Sigma-Algebra

A sigma-algebra is a collection of subsets, including the entire set and the empty set, that is closed under complementation and countable unions. In the context of locally measurable sets, $\tilde{\mathcal{M}}$ forms a sigma-algebra. This is crucial because it accommodates operations like taking complements, and forming countable unions while retaining measurability.

• Closure under complementation ensures if a set is measurable, so is its complement within the defined measure space.
• Closure under countable unions allows for the construction of larger sets from smaller measurable ones.
• It always contains the full space $X$ and the empty set $\mathrm{\varnothing }$.

By proving $\tilde{\mathcal{M}}$ is a sigma-algebra, we validate it as a robust framework for dealing with measure theory challenges.

Complete Measure

A measure $\mu$ is complete if every subset of a measurable set with measure zero is also measurable. Completeness is about ensuring that tiny (or negligible) pieces of the space don't disrupt the measurability of the entire space.

When extending a measure to include locally measurable sets (as with $\tilde{\mu }$), this completeness is maintained if $\mu$ was complete. This means that any set, or part of a set, that was negligible before remains so, preserving the structure and integrity of the measure space. This feature is vital when dealing with real-world applications where null sets can often arise, and it ensures that the entirety of the measure space is well-delineated and usable.

Semifinite Measure

A semifinite measure is a measure where every set of infinite measure contains a subset with finite, non-zero measure. This feature prevents the possibility of having entire regions of a space that are immeasurable and ensures that every part of the space can be measured to some extent. This condition is especially helpful in infinite measure spaces.

When constructing a saturated measure from a semifinite measure, we determine the measure for locally measurable sets by taking the supremum of the measures of the subsets. This ensures that measurements are comprehensive and extend the original measure $\mu$ in a meaningful way. By handling infinite sets effectively, semifinite measures anchor the measure space's robustness, even if it otherwise consists of complex or vast regions.