Indicator functions are independent I'm trying to prove that :

Suppose $(\Omega, \mathcal{F}, \mathbb{P})$ is a probability space and $A,B \in \mathcal{F}$.
Then $1_A, 1_B$ are independent random variables if and only if $A, B$ are independent events.

Using the following definition of independence of random variables (Measure Integration and Real Analysis, Axler 2020):

Two random variables $X, Y$ are called independent if ${X \in U}$ and ${Y \in V}$ are independent events for all Borel sets $U, V \in \mathbb{R}$

But I can't connect the two probabilities $P(A)$ and $P(1_A \in U)$ with $U$ is a Borel set in $\mathbb{R}$.
Could you please give some suggestions on this proof ?
 A: Notice that $P(1_A =1)=P(A), P(1_A =0)=P(A^c)$.
Same goes for $B$.
Then $P(1_A=1,1_B=1)= P(A \cap B)=P(A)P(B) = P(1_A=1)P(1_B=1)$ iff $A$ and $B$ are independent.
We treated the case when $1_A=1$ and $1_B=1$. There are three more cases to treat. Once you treat those, you will have showed that the mass function of the vector $(1_A,1_B)$ factorises, which implies that your definition of independence is satisfied for Borel sets of the form $]-\infty,x]$ for all $x\in \mathbb R$. But those sets generate the Borel $\sigma$-algebra.
A: $\implies$
$$
\mathbb{P}(\{\mathbf{1_A}=1\}\cap\{\mathbf{1_B}=1\})
=\mathbb{P}(\{\mathbf{1_{A\cap B}}=1\})
=\mathbb{P}(A\cap B)
$$
By definition of independent r.v, those corresponding signed events need to be independent.
$$\mathbb{P}(\{\mathbf{1_A}=1\}\cap\{\mathbf{1_B}=1\})=\mathbb{P}(\{\mathbf{1_A}=1\})\times\mathbb{P}(\{\mathbf{1_B}=1\})=\mathbb{P}(A)\mathbb{P}(B)$$
So that we obtained  $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$
$\impliedby$
Given $$\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)\\
\mathbb{P}(\{\mathbf{1_{A\cap B}}=1\})=\mathbb{P}(\{\mathbf{1_A}=1\})\mathbb{P}(\{\mathbf{1_B}=1\})\\
\mathbb{P}(\{\mathbf{1_A1_B}=1\})=\mathbb{P}(\{\mathbf{1_A}=1\})\mathbb{P}(\{\mathbf{1_B}=1\})\\
\mathbb{P}(\{\mathbf{1_{A}}=1\}\cap\{\mathbf{1_B}=1\})=\mathbb{P}(\{\mathbf{1_A=1}\})\mathbb{P}(\{\mathbf{1_B=1}\})
$$
so that $\mathbf{1_A}$ and $\mathbf{1_B}$ are independent r.v
