note, video

If $A \In B$ are both measurable, then $B \RM A$ is measurable, and $m(B \RM A) = m(B) - m(A)$.

We need to show that for any subset $E \In \R^n$… wait a second, do we really need to go by definitions again? After all these preparations, we should be able to exploit our sweat. Hint

• $B \RM A = B \cap A^c$.
• $B = A \sqcup (B \RM A)$, a decomposition into measurable subsets.

Let $\{E_j\}_{j=1}^\infty$ be a countable collection of disjoint subsets. Then, their union $E$ is measurable, and we have

$$ m^*(E) = \sum_{j=1}^\infty m^*(E_j) $$

Proof: Tao's proof is really clever, let's first try to go through the proof, then discuss how we can come up with it ourselves.

First, we start by showing $E$ is measurable from the definition: we want to show for any subset $A$, we have $$ m^*(A) = m^*(A \cap E) + m^*(A \RM E) $$ Suffice to show $\geq$ direction. Let $F_N = \sum_{j=1}^N E_j$. We have two expressions

• $m^*(A \cap E) \leq \sum_{j=1}^\infty m(A \cap E_j) = \sup_{N > 1} \sum_{j=1}^N m^*(A \cap E_j) = \sup_{N > 1} m^*(A \cap F_N) $
• $m^*(A \RM E) \leq m^*(A \RM F_N)$ for all $N$

Hence, $$ m^*(A \RM E) + m^*(A \cap E) \leq \sup_{N > 1} (m^*(A \cap F_N)) + m^*(A \RM E)) = \sup_{N > 1} (m^*(A \cap F_N)) + m^*(A \RM F_N)) = \sup_{N > 1} m^*(A) = m^*(A) $$

OK, that shows $E$ is measurable. To finish off, we need to show countable addivity $$ m^*(E) = \sum_{j=1}^\infty m^*(E_j) $$ Since $E = \cup E_j$, we have $\leq$ from countable sub-addivity. Then, by monotonicity, we have $$ m^*(E) \geq m^*(F_N) = \sum_{j=1}^N m^*(E_j) $$ since this is true for all $N$, we can sup over $N$, and get $$ m^*(E) \geq \sup_N \sum_{j=1}^N m^*(E_j) = \sum_{j=1}^\infty m^*(E_j) $$

One slogan is to approximate $E$ by $F_N$. We want to prove $$ m^*(A) \geq m^*(A \cap E) + m^*(A \RM E) $$ If we have

• (1) $m^*(A \cap E) \leq m^*(A \cap F_N)$ and
• (2) $m^*(A \RM E) \leq m^*(A \RM F_N)$,

then we can write $$ m^*(A \cap E) + m^*(A \RM E) \leq m^*(A \cap F_N) + m^*(A \RM F_N) = m^*(A) $$ but unfortunately, (1) is wrong. One way to remedy this, is to show that (assuming $m^*(A)< \infty$), for any $\epsilon > 0$, there exists an $N$, such that $m^*(A \cap E) \leq m^*(A \cap F_N) + \epsilon$ holds. (see if you can make this approach work). Another more elegant approach is done as above, using countable subadditivity to get $\leq$, then introduce a $\sup$ to get to finite $N$.

Try to forget this proof, and come up with your own. It might be fun.

The $\sigma$-algebra property.

Given a countable collection of measurable set $\Omega_j$, one need to prove that $\cup \Omega_j$ and $\cap \Omega_j$ are measurable.

We only need to prove the case of $\Omega = \cup_j \Omega_j$, since the $\cap$ operation can be obtained by taking complement and $\cup$. The hint is to define $$ \Omega_N = \cup_{j=1}^N \Omega_j$$ and $E_N = \Omega_N \RM \Omega_{N-1}$, then $\{E_j\}$ are measurable, mutually disjoint, and $\cup_j E_j = \cup_j \Omega_j = \Omega$.

Every open set can be written as a finite or countable union of open boxes.

I will leave this as discussion problem.

• A subset $U$ is open, if for every point $x \in U$, there exists an open ball $B(x,r) \In U$.
• claim: a subset $U$ is open, iff $\forall x\in U$, there exists an open box $B$, such that $x \in B \In U$.
• claim: a subset $U$ is open, iff $\forall x\in U$, there exists an open box $B$ with rational boundary coordinates, such that $x \in B \In U$.
• There are countably many open boxes with rational boundary coordinates.

All open sets are measurable.

Since open boxes are measurable, and countable union of measurable sets are measurable.

An alternative definition for measurable set is the following:

A subset $E$ is measurable, if for any $\epsilon>0$, there exists an open set $U\supset E$, such that $m^*(U \RM E) < \epsilon$.

Can you show that this definition is equivalent to the Caratheodory criterion (the one we had been using)?