Csdrhrt

Let $E$ be the union of a countable collection of measurable sets.

Then there is a countable disjoint collection of measurable sets {$E_k$}$_{k=1}^infty$ for which $E = cup_{k=1}^infty E_k$.

Let $A$ be any set. Let $n$ be a natural number.

Define $F_n = cup_{k=1}^n E_k$.

Since $F_n$ is measurable and $F_n^{c}supset E^c$,

$m^*(A) = m^*(Acap F_n) + m^*(Acap F_n^c) ge m^*(Acap F_n) + m^*(Acap E^c)$ , and

$m^*(Acap F_n) = sum_{k=1}^infty m^*(Acap E_k)$. (Since {$E_k$}$_{k=1}^n$ is a finite disjoint collection of measurable sets.)

Thus $m^*(A)ge sum_{k=1}^n m^*(Acap E_k) + m^*(Acap E^c) $

The left-hand side of this inequality is "independent" of n.....(omitted)

I want to ask what "indepdendent" means? and why it is "independent" of n?

We can say that it is independent of $n$ in the following way, following this reference.

By additivity of finite unions, we have: $m^*(A cap F_n) = m^*(A cap E_1) + ldots m^*(A cap E_n)$ for arbitrary $n$.

If we let $n rightarrow infty$, then the LHS tends to $m^*(A cap E)$ and the RHS tends to $sum_{k=1}^infty m^*(A cap E_k)$. Since the finite sum is composed of positive numbers, if we consider the full expression $m^*(A) geq sum_{k=1}^n m^*(A cap E_k) + m^*(A cap E^C)$, for each $n$, we see that the partial sum is bounded. This means that the infinite sum is bounded, and hence "independent" of $n$.

Finally, we can use the subadditivity of of outer measure on the term $sum_{k=1}^infty m^*(A cap E_k)$ and see that it is larger than the union of the $(A cap E_k)$ terms, i.e., $sum_{k=1}^infty m^*(A cap E_k) geq m^* bigcup_{k=1}^infty (A cap E_k)$.

So my takeaway is that "independent" means that we can make this argument for any $n$, and as $n$ approaches infinity, the union becomes $E$, i.e., the leftover "stuff" in $E$ that is not yet included in the union is measure zero (thought I'm not sure if this interpretation is right).

Suppose you change the line "Let $A$ be any set. Let $n$ be a natural number." to "Let $A$ be any set. Then $forall nin Bbb N....$"

The inequality you have difficulty with should then say $forall nin Bbb N;( m^*(A)geq$... (et cetera)...). The last inequality holds for $every $ $nin Bbb N$

It follows immediately that $m^*(A)geq$ $ m^*(Acap E^c)+sup_{nin Bbb N}sum_{k=1}^nm^*(Acap E_k)=$ $=m^*(Acap E^c)+sum_{k=1}^{infty}m^*(E_k).$