$\forall \alpha \in [0,1] \exists B_{\alpha} \in \mathcal B(\Bbb R): \Bbb P(B_{\alpha}) = \alpha$

Question

Let $\Bbb P$ be a probability measure on $\Bbb R$ with $\Bbb P(\{x\}) = 0$ for all $x \in \Bbb R$. I want to show that for all $\alpha \in [0,1]$ there exists a $B_{\alpha} \in \mathcal B(\Bbb R)$ such that $\Bbb P(B_{\alpha}) = \alpha$.

My approach is to take an interval $I_m = [-m,m]$ (let's assume $\Bbb P([-m,m]) > \alpha$) and construct a sequence of subintervals $[-m + q_k, m - q'_k]$ with $(q_k,q'_k) \in \Bbb Q_+^2$ such that $[-m + q_k, m - q'_k] \supseteq [-m + q_{k+1}, m - q'_{k+1}]$ and $\Bbb P ([-m + q_k, m - q'_k]) \ge \alpha $ for all $k \in \Bbb N$. Then my idea was to use the continuity from above of the measure but I'm not sure whether this sequence indeed converges to a set with measure $\alpha$.

Answer 1

Let $F(x)=\mathbb{P}((-\infty,x])$ be the c.d.f. associated with $\mathbb{P}$. The condition $\mathbb{P}(\{x\})=0$ for each $x\in\mathbb{R}$ implies that $F$ is a continuous function. Note that $\lim_{x\rightarrow-\infty}F(x)=0$ and $\lim_{x\rightarrow\infty}F(x)=1$. By intermediate value theorem, the result follows.

Answer 2

The cumulative density function is continuous iff $\mathbb{P}(\{x\})=0$ for all $x$. As this applies, we have a continuous $F$ with limit 0 at $-\infty$ and with limit 1 at $\infty$. So the range of $F$ is $[0,1]$, making your assertion true.