Let $\delta_x$ be the dirac measure and $\delta_x^*$ the corresponding outer measure. Show that for all $A \subset (0,1]$ the equation
$\delta_x^*(A):=\cases{1, x \in A \\ 0, x \notin A}$
holds.
Note that the outer measure is definded as
$\mu^∗
(A) = \inf \{\sum_{j=1}^\infty
\mu(A_j ) : A_j ∈ B, j ∈ \mathbb{N},
\bigcup_{j=1}^\infty
A_j ⊃ A\}$
In this case B is the ring of all finite unions of left-half-open intervals.
What I've tried:
Define the set $\{x:x \in A\}$ and show that the outer measure on it is $1$. Let $(A_n)$ be a sequence in $B$. Then this set is a subset of $\bigcup_{n \in \mathbb{N}}A_n.$ Hence there has to be a $m \in \mathbb{N}$ so that $x \in A_m$.
I'm stuck on this point. I also need to prove that all subsets $A$ of $(0,1]$ are $\delta_x^*$-measurable. Any help is appreciated!
Best Answer
Let $A\subseteq(0,1]$. It is easy to see that
$$0\leq\delta_x^*(A)\leq1.$$
Suppose first that $x\in A$. Then for any sequence $\{A_j\}_{j\in\mathbb{N}}$ in $B$ with $A\subseteq\bigcup_{j\in\mathbb{N}}A_j$ we have that $x\in A_k$ for some $k\in\mathbb{N}$. In particular then
$$\sum_{j=0}^\infty\delta_x(A_j)\geq\delta_x(A_k)=1.$$
Taking the infimum over all such sequences $\{A_j\}_{j\in\mathbb{N}}$ we get that $\delta_x^*(A)\geq1$, and so $\delta_x^*(A)=1$.
Suppose now that $x\notin A$. Choose a sequence $\{A_j\}_{j\in\mathbb{N}}$ in $B$ such that $A\subseteq\bigcup_{j\in\mathbb{N}}A_j=(0,1]\setminus\{x\}$ (this can easily be done explicitly if you're unsure about why such a sequence exists). This means that $\delta_x(A_j)=0$ for all $j\in\mathbb{N}$, and so
$$0\leq\delta_x^*(A)\leq\sum_{j=0}^\infty\delta_x(A_j)=0,$$
from which it follows that $\delta_x^*(A)=0$.
So we have now proven than $\delta_x^*(A)=\delta_x(A)$ for any $A\subseteq(0,1]$. I'll leave the measurability part of the question to you, as that should be a very short and straightforward argument.