continuity of a probability measure if $\mu (\{x\})=0$

Question

Let $\mu$ a probabibility measure in $\mathbb R$ with $\mu\{x\}=0\quad \forall x\in\mathbb R$. Show that for all $\alpha \in [0,1]$ there exists $B_\alpha \in\mathcal B(\mathbb R)$ with $\alpha=\mu (B_\alpha)$.

My thoughts: We have to show that the distribution function $F(y)=\mu(-\infty,y)$ is continous. Then we can apply the intermediate value theorem.

I am not sure how to show that this measure is continous. Some help is welcome!

Answer 1

The probability measure $\mu$ is continuous from above in the sense that for any sequence $A_1 \supseteq A_2 \supseteq \dots$ of measurable sets it holds that

$$\mu \left( \bigcap_{n \in \mathbb{N}} A_n \right) = \lim_{n \to \infty} \mu(A_n).$$

To prove right-continuity of the distribution function, choose an arbitrary sequence $x_n \downarrow x$ and consider the sets $A_n := (-\infty,x_n)$ to show that

$$\mu((-\infty,x]) = \lim_{n \to \infty} \mu(A_n) = \lim_{n \to \infty} \mu((-\infty,x_n)).$$

Finally, you have to use the fact that $\mu(\{x\})=0$ to conclude that $F$ is right-continuous at $x$.

Answer 2

Use what is called "continuity of the measure": If $(A_n)$ is a decreasing sequence of measurable sets of finite measure, then $\mu(\bigcap_n A_n) =\lim_n \mu(A_n)$.

Now apply this with $A_n =[x_0, x_0 +1/n]$. Depending on what exactly you need, you might also take $A_n =[x_0-1/n, x_0]$, or $A_n = [x_0-1/n,x_0+1/n]$.