Exercise
Let $m$ be a Borel probability measure on $[0,1]$ and let $m\otimes m$ be the product measure on $[0,1]^2$.
(a) Show that exists a sequence of points $\{x_n\}_{n \in \mathbb{N}} \subset [0,1]$ and a Borel measure $m^c$ on $[0,1]$ s.t. $$ m = \sum_{n \in \mathbb{N}} c_n \delta_{x_n} + m^c $$ where $c_n := m(\{x_n\})$, $\delta_x$ is the Dirac measure at $x$ and $m^c(\{x\})=0$ for all $x \in [0,1]$.
(b) Let $D:=\{(x,y) :x=y \} \subset [0,1]^2$. Prove $$ m \otimes m (D) = \sum_{n \in \mathbb{N}} c_n^2$$
(c) Prove $$ \sup\{m(\{x\}), x \in [0,1] \} \ge m \otimes m (D) $$
Attempt
(a) Let $X:= \{x \in [0,1] \, | \, m(\{x\}) >0 \}$. Suppose $X$ is uncountable. In any case $$X = \bigcup_{n \in \mathbb{N}_0} \{x \in [0,1] \, | \, m(\{x\}) \ge 1/n \} = \bigcup_{n \in \mathbb{N}_0} X_n$$ Since $X$ is uncountable there must be $N \in \mathbb{N}_0$ s.t. $X_N$ is infinite. Take $J \subset X_N$ s.t. $|J|=3N$. Then $$ m([0,1]) \ge m(J) \ge 3N \frac{1}{N} = 3 $$ wich is a contradiction. Then $X$ is countable and set $\{x_n\}_{n \in \mathbb{N}}:=X$.
(b) Clearly $m \otimes m (\{ (x_n,x_n)\}_{n \in \mathbb{N}}) = \sum_{n \in \mathbb{N}} c_n^2$.
Set $C=D \setminus \{ (x_n,x_n)\}_{n \in \mathbb{N}}$. I want to prove that $m \otimes m(C)=0$. Let $a:=m^c([0,1])$.
First I show the following: For all $N \in \mathbb{N}_0$ exists a disjoint finite partition $\{A_1, \dots, A_N \}$ of $[0,1]$ s.t. $m^c(A_i) = a/N$ for all $i=1, \dots, N$. Indeed, take $x_1 \in [0,1]$, and $f_1(t) = m^c((x_1-t,x_1+t)\cap[0,1])$. Then $f_1$ is continuous and $f_1(1)=a \ge f_1(0)=0$. Then $$ \exists t_1 \in [0,1] \text{ s.t. } f_1(t_1)=a/N$$ Then take $A_1 :=(x_1-t_1,x_1+t_1)\cap[0,1]$. Then consider $x_2 \in [0,1] \setminus A_1$ and $f_2(t) = m^c((x_2-t,x_2+t)\cap([0,1]\setminus A_1))$ and so on.
This procedure can be iterated again $N-1$ times constructing all the $A_i$.
Now, fix $\epsilon>0$. Let $N:= \left \lceil {a^2/\epsilon} \right \rceil$. Take a partition $\{A_1, \dots, A_N \}$ as described above. Set $B_i := A_i \setminus \{x_n\}_{n \in \mathbb{N}}$ for all $i=1, \dots, N$. Then
$$m \otimes m (C) \le m \otimes m \Biggl( \bigcup_{i=1}^{N} B_i^2 \Biggr) = \sum_{i=1}^{N} m(B_i)^2 =\sum_{i=1}^{N} m^c(A_i)^2 =a^2/N \le \epsilon $$
Then $m \otimes m (C)=0$.
(c) Clearly $\sup\{m(\{x\}), x \in [0,1] \} = \sup_{n \in \mathbb{N}} c_n$ and $\sum_{n \in \mathbb{N}} c_n \le m([0,1])=1$. Then
$$ m \otimes m (D) =\sum_{n \in \mathbb{N}} c_n^2 \le \sup\{m(\{x\}), x \in [0,1] \} \sum_{n \in \mathbb{N}} c_n \le \sup\{m(\{x\}), x \in [0,1] \} $$
Doubts
I'm quite sure about the proof but it is really long. I would really appreciate your advices and opinions!