Question

The inner measure ${\mu }_{\ast }(E)$ of a set $E$ is defined as the least upper bound of the measures of all measurable subsets of $E$. Show that ${\mu }_{\ast }(E)\le {\mu }^{\ast }(E)$. For any open set $U\supset E$, show that $$\mu (U)={\mu }_{\ast }(U\cap E)+{\mu }^{\ast }(U-E)$$ and that $E$ is measurable with finite measure if and only if ${\mu }_{\ast }(E)={\mu }^{\ast }(E)<\mathrm{\infty }$.

Short answer

The inequality ${\mu }_{\ast }(E)\le {\mu }^{\ast }(E)$ holds due to the definitions of inner and outer measures. For an open set $U$ containing $E$, $\mu (U)={\mu }_{\ast }(U\cap E)+{\mu }^{\ast }(U-E)$. A set $E$ is measurable with finite measure if and only if ${\mu }_{\ast }(E)={\mu }^{\ast }(E)<\mathrm{\infty }$.

Step-by-step solution

Proof of Inequality

According to the definitions of inner and outer measure, it is clear that ${\mu }_{\ast }(E)\le {\mu }^{\ast }(E)$. The inner measure of $E$ is the supremum of the measures of all measurable subsets of $E$. So it cannot exceed the infimum of all measures of the open sets containing $E$ that is the outer measure.

Proof of Equality for Open Set

We need to prove $\mu (U)={\mu }_{\ast }(U\cap E)+{\mu }^{\ast }(U-E)$ for any open set $U$ that contains $E$. By subadditivity of measures, we get $\mu (U)\le \mu (U\cap E)+\mu (U-E)$. Since ${\mu }_{\ast }(U\cap E)\le \mu (U\cap E)$ and ${\mu }^{\ast }(U-E)\le \mu (U-E)$, summing up these inequalities, we get ${\mu }_{\ast }(U\cap E)+{\mu }^{\ast }(U-E)\le \mu (U)$. Hence proved that $\mu (U)={\mu }_{\ast }(U\cap E)+{\mu }^{\ast }(U-E)$ for any open set $U$ that contains $E$.

Measurability and Finiteness of $E$

$E$ is measurable with finite measure if and only if ${\mu }_{\ast }(E)={\mu }^{\ast }(E)<\mathrm{\infty }$. By definition, a set is measurable if the inner and outer measures are equal, and finite if the measure is not infinite. Hence the result follows directly from the definition.

Key concepts

Inner Measure

The concept of an inner measure is fundamental in measure theory. We define the inner measure of a set $E$, denoted as ${\mu }_{\ast }(E)$, as the least upper bound (supremum) of the measure of all its measurable subsets. This might sound complicated, but let's break it down.

Imagine that we are trying to "fill" the set $E$ with the largest possible measurable subsets. Each subset contributes to a total measure, and ${\mu }_{\ast }(E)$ is the largest such value that can be achieved without exceeding the actual size of $E$.

Inner measure allows us to understand the "inside" of a set in terms of its measurable components. It's an essential tool for capturing more nuanced details about the structure and size of sets, particularly when dealing with sets that are not easily measurable in the traditional sense. This measure provides an internal view of $E$, focusing on what can be neatly packed within it by measurable sets.

Outer Measure

Outer measure, denoted ${\mu }^{\ast }(E)$, is another crucial concept in measure theory. It helps us understand the size of a set from an "external" perspective. To determine the outer measure of a set $E$, we look for the smallest possible total measure of open sets covering $E$.

Think of covering $E$ with an open 'wrap'. These wraps are open sets, and among all possible ways to cover $E$, we choose the one with the smallest measure. This smallest or infimum measure is what we call the outer measure of $E$.

The outer measure is important because it captures the minimal size necessary to envelop $E$. It's a bound on the external environment of the set, hence providing an alternative angle to measure sets. In practice, comparing inner and outer measures helps determine measurability.

Measurable Sets

A set $E$ is deemed measurable when its inner measure equals its outer measure. In simpler terms, if ${\mu }_{\ast }(E)={\mu }^{\ast }(E)$, the set $E$ can be measured seamlessly, without any ambiguity between its interior and exterior.

For sets with finite measure, this equality is particularly significant. It indicates that the set is well-defined and can be measured uniformly regardless of view (from inside using inner measure or outside using outer measure).

Measurable sets are vital in measure theory because they ensure consistency and reliability in size determination. Having ${\mu }_{\ast }(E)={\mu }^{\ast }(E)<\mathrm{\infty }$ means that $E$ can be dealt with using conventional measure techniques, which simplifies analysis and computation.

Subadditivity of Measures

Subadditivity is an important property that all measures possess. It highlights that the measure of a union of two sets is less than or equal to the sum of their individual measures.

Mathematically, if $A$ and $B$ are any two sets, their measure satisfies $\mu (A\cup B)\le \mu (A)+\mu (B)$. This is called subadditivity, and it reflects the intuitive notion that the total size should not be more than the sum of the parts.

Subadditivity becomes instrumental in proving equalities like $\mu (U)={\mu }_{\ast }(U\cap E)+{\mu }^{\ast }(U-E)$ for open sets $U$ containing $E$. By using subadditivity, we establish bounds that facilitate an equality relating inner and outer measures, which is essential for demonstrating properties of measurable and non-measurable sets.