Dgdrxrt

Why the definition of the outer measure $mu^*$:

$$mu^*(A) = infBiglsum_n=1^infty mu(A_n): A subset bigcup_n=1^infty A_n, A_n in mathcal A,; n in mathbb NBigr$$

is wrong when $A_i cap A_j not= emptyset$, $i not = j$?

Can someone help me to show that?

I'm not quite sure what you are asking, but the the following may be insightful. Suppose I also define,

$$barmu^*(A) = infBiglsum_n=1^infty mu(A_n): A subset bigcup_n=1^infty A_n, A_n in mathcal A; textare pairwise disjoint Bigr$$

I claim that $mu^* = bar mu^*.$

Since we are taking the infimum over a smaller collection of sets,we have $mu^*(A) leq barmu^*(A)$ for all $A subset X$ (here $X$ is your measure space).

Conversely let $A_n in mathcal A$ be a sequence such that $A subset bigcup_n=1^infty A_n.$ Then we define $B_n in mathcal A$ as $B_1 = A_1$ and for $n geq 2,$
$$ B_n+1 = A_n+1 setminus bigcup_i=1^n A_i.$$
Observe by construction that $B_n$ are pairwise disjoint and we have,
$$ bigcup_n=1^infty A_n = bigcup_n=1^infty B_n. $$
Therefore we have,
$$ barmu^*(A) leq sum_n=1^infty mu(B_n) leq sum_n=1^infty mu(A_n). $$
Taking the infimum over all sequences $A_n in mathcal A$ we deduce that,
$$ barmu^*(A) leq mu^*(A). $$
Hence they are equal.

Although these two definitions are the same, the reason we usually don't take pairwise disjoint sequences when defining the outer measure is because it tends to be simpler to work with. If we have a covering of our set $A$ we want to compute the outer measure of, we don't need to go through this construction to get a disjoint covering each time.

Also in practice the technical construction of the outer measure is not too important, beyond the fact that it can be constructed. The important point is that it satisfies certain properties as outlined on the wikipedia page, and the provided definition is the minimal thing that works.