Assume $ F $ is a non-decreasing, right continuous function $ F: \mathbb{R} \to [0,1] $ such that
$F(\infty)=1 $ and $F(-\infty) = 0 $.
By Lebesgue Stieltjas measure theorem, there exists a unique probability measure denoted by $ dF $, such that $ (\mathbb{R},\mathcal(B),dF) $ is a probability space.
Now let for any $ D \in \mathcal{B} $ , let $ \mathcal{B} (D) = \{A\cap D :A\in \mathcal{B}\}$.
For any $ A\in \mathcal{B} $ such that $ A\subseteq [0,1] $ define $ \mathbb{P}(A)=dF(A) $.
Is it true that $ ([0,1],\mathcal{B}([0,1]),\mathbb{P} $ is a probability space?
Best Answer
No, not in general, since you cannot conclude that $dF([0,1])=1$. You can actually compute the total mass as $$dF([0,1]) = F(1) - \lim_{x\uparrow 0}F(x)$$ and thus $\mathbb{P}$ is a probability measure if and only if $F(1)=1$ and $\lim_{x\uparrow 0}F(x)=0$, where the last condition is equivalent to $F(0)=0$.
If you want to secure that $\mathbb{P}$ becomes a probability measure, you can instead define it as $$\mathbb{P}(A) = \frac{dF(A)}{dF([0,1])} \quad \quad (A\subseteq [0,1]).$$ The last formula should look familiar and can be interpreted as the conditional probability of the event $A$ given that we know that the outcome is inside $[0,1]$.