# Measure Theory - Existence of Fsigma set contained in measurable set

• MHB
• joypav
In summary, we have shown that if $E$ has finite outer measure, then $E$ is measurable if and only if there is a $F_\sigma$ set $F \subset E$ with $m^*\left(F\right)=m^*\left(E\right)$. This can be proven in both directions by using the representation $E=K\cup N$ where $K$ is $F_\sigma$ and $m^*(N)=0$.
joypav
Problem:
Let $E$ have finite outer measure. Show that $E$ is measurable if and only if there is a $F_\sigma$ set $F \subset E$ with $m^*\left(F\right)=m^*\left(E\right)$.

Proof:
"$\leftarrow$"
To Show: $E=K\cup N$ where $K$ is $F_\sigma$ and $m^*(N)=m(N)=0$.

By assumption, $\exists F$, and $F_\sigma$ set, $F\subset E$, and $m^*(F)=m^*(E)$
Write,
$E=F\cup (E-F)$

To Show: $m^*(E-F)=0$

By a theorem given in class,
$\exists G$, a $G_\delta$ set, such that $E\subset G$ and $m^*(E)=m(G)$.
Then,
$m(E-F) \leq$ (G is "bigger" than E) $m(G-F) = m(G)-m(F) = m(E) - m(E) = 0$
$\implies m^*(E-F)=m(E-F)=0 \implies E$ is measurable.

Could someone get me started in the other direction??

If $E$ is measurable, use the representation you wrote above: $E = K\cup N$ where $K$ is $F_\sigma$ and $m^*(N) = 0$. By countable subadditivity of the outer measure $m^*(E) \le m^*(K) + m^*(N) = m^*(K)$. On the other hand, since $K \subset E$, $m^*(K) \le m^*(E)$ by monotonicity of $m^*$. Therefore, $m^*(E) = m^*(K)$.

"$\rightarrow$"
To Show: $m^*(F)=m^*(E)$.

Assume $E$ is measurable. Then, by definition, there exists a $F_\sigma$ set $K$ such that $E=K\cup N$ where $m^*(N)=0$.

Since $K$ is $F_\sigma$, it can be written as a countable union of closed sets, $K=\bigcup_{i=1}^\infty F_i$.

Then, by countable subadditivity of outer measure,
$m^*(E) = m^*(K\cup N) \leq m^*(K) + m^*(N) = \sum_{i=1}^\infty m^*(F_i) + 0$.

But since $F_i \subset K \subset E$, we have $m^*(F_i) \leq m^*(E)$ for all $i$, so
$\sum_{i=1}^\infty m^*(F_i) \leq \sum_{i=1}^\infty m^*(E) = \infty$.

Therefore, $m^*(E) \leq \infty$, and since $E$ has finite outer measure, we must have $m^*(E) < \infty$.

Now, by monotonicity of outer measure, we have $m^*(F_i) \leq m^*(E)$ for all $i$, so
$\sum_{i=1}^\infty m^*(F_i) \leq \sum_{i=1}^\infty m^*(E) = m^*(E) < \infty$.

Since the series converges, we can rearrange the terms to get
$\sum_{i=1}^\infty m^*(F_i) = m^*(E) - m^*(N)$.

But since $m^*(N)=0$, we have
$\sum_{i=1}^\infty m^*(F_i) = m^*(E)$.

Therefore, $m^*(F)=m^*(K)=m^*(E)$, as desired.

## 1. What is the definition of an Fsigma set?

An Fsigma set is a countable union of closed sets. In other words, it is a set that can be written as the union of countably many closed sets.

## 2. How does the existence of an Fsigma set contained in a measurable set relate to measure theory?

In measure theory, the existence of an Fsigma set contained in a measurable set is important because it allows us to approximate the measurable set by simpler sets. This is useful in many mathematical proofs and applications.

## 3. Can an Fsigma set be uncountable?

Yes, an Fsigma set can be uncountable. For example, the set of all irrational numbers can be written as the countable union of singletons (which are closed sets), making it an uncountable Fsigma set.

## 4. How do you prove the existence of an Fsigma set contained in a measurable set?

The existence of an Fsigma set contained in a measurable set can be proven using the Carathéodory's Extension Theorem. This theorem states that any outer measure on a measurable set can be extended to a measure on the sigma-algebra generated by that set.

## 5. Why is the existence of an Fsigma set contained in a measurable set important?

The existence of an Fsigma set contained in a measurable set is important because it allows us to approximate the measurable set by simpler sets, making it easier to work with in mathematical proofs and applications. Additionally, it is a fundamental concept in measure theory and is used in many other areas of mathematics.

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