Showing that $\mathbb P$ is a measure Let $\Omega=\{\omega=(y_1,_\ldots,y_T):y_i=\pm1\}$
Define $\mathbb P(\{\omega\})=x^{k(\omega)}(1-x^{T-k(\omega)})$
where $x:=\frac{r-a}{b-a}, a<r<b$ and $k(\omega)$ counts the number of $1$ in $\omega$.
I need to show that  $\mathbb P$ defines a probability measure without using the product structure of $\Omega=\{-1,1\}^T$.

I could already show that $\mathbb P\ge0$ and $\mathbb P(\Omega)=1$ but I'm struggling with additivity, so  I would appreciate it if someone could help me to show 

$$\mathbb P(\{\omega_i,\omega_j\})=\mathbb P(\{\omega_i\})+\mathbb P(\{\omega_j\})\quad \textrm{for}\ i \neq j$$
 A: In your situation it can be proved that $\mathbb P$ induces a probability measure on measurable space $(\Omega,\wp(\Omega))$.
For this define an extension of $\mathbb P$ by stating that by definition:$$\mathbb P(A)=\sum_{\omega\in A}\mathbb P(\{\omega\})$$This definition is okay since $\Omega$ is a countable set, and further it is evident that it extends the original $\mathbb P$ which is only defined on singletons.
After that it it must be checked that $\mathbb P(A)\geq0$ for every $A$ (which follows directly from the fact that $\mathbb P$ is non-negative on singletons) and that $\mathbb P(\Omega)=1$.
Actually the check that $\mathbb P(\Omega)=1$ is the main thing, and additivity is already assured now. This on base of a more general observation that I will give you under the line.

If $S$ is a countable set and $p_s\in[0,\infty)$ for every $s\in S$ such that $\sum_{s\in S}p_s=1$ then the function $P:\wp(S)\to\mathbb R$ prescribed by: $$A\mapsto\sum_{s\in A}p_s$$is a probability measure.
If $(A_n)_n$ are disjoint subsets of $S$ for $n=1,2,\dots$ then:$$P\left(\bigcup_{n=1}^{\infty}A_n\right)=\sum_{s\in\bigcup_{n=1}^{\infty}A_n}p_s=\sum_{s\in S}p_s\mathbf1_{\bigcup_{n=1}^{\infty}A_n}(s)=\sum_{s\in S}p_s\sum_{n=1}^{\infty}\mathbf1_{A_n}(s)=\sum_{n=1}^{\infty}\sum_{s\in S}p\mathbf1_{A_n}(s)=\sum_{n=1}^{\infty}P(A_n)$$
Underlying is the fact that for nonnegative constants $c_{n,m}$ it is true that:$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}c_{n,m}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}c_{n,m}$$
