# Thread: Probability measure on Borel-sigma

1. ## Probability measure on Borel-sigma

Let $\displaystyle (\Omega , \mathcal{F} , \mathbb{P} )$ be a probabilty space and $\displaystyle X: \ \Omega \rightarrow \ \bar{\mathbb{R}}$ be a random variable with distribution $\displaystyle \mu$. Show that $\displaystyle \mu$ is a probabilty measure on $\displaystyle \bar{\mathcal{B}}$. That is, verify the three axioms of probabilty for $\displaystyle \mu$, using the collection of "events" on $\displaystyle \bar{\mathcal{B}}$.

2. Hello,

But how is $\displaystyle \mathcal{\overline B}$ defined ?
Is it the Borel algebra over $\displaystyle \mathbb{\overline R}$ ?

3. Hi Moo,
Yes, $\displaystyle \mathcal{\overline B}$ is the borel algebra over $\displaystyle \mathbb{\overline R}$

4. I've become stuck with the question. My teacher said that I need to use the fact that $\displaystyle \mathbb{P}$ is a probability measure and that $\displaystyle X$ is a random variable. I've racked my brain but can't work it out Whats more annoying is that the answer is staring me right in the face.
I know that the 3 axioms of probability are

(1) $\displaystyle 0 \leq \mathbb{P} \leq 1$ for some event A

(2) $\displaystyle \mathbb{P} (\Sigma) = 1$

(3) For any sequence $\displaystyle A_1, A_2, ...$ which are disjoint
$\displaystyle \mathbb{P} ( \bigcup_i A_i ) = \sum_i \mathbb{P} (A_i).$

I just can't get my head around this and any help would be greatly appreciated.

5. I'm just learning about probability and measure theory myself, and I am not a mathematician. So here's my best shot at it:

We have that $\displaystyle X: (\Omega,\mathcal{F}) \rightarrow (\mathbb{\bar{R}},\mathcal{\bar{B}})$, and I suppose that makes $\displaystyle X$ an extended random variable. Now, $\displaystyle X$ induces a probability measure on $\displaystyle \bar{\mathcal{B}}$, namely$\displaystyle \mathbb{P}_X(B) = \mathbb{P}\{\omega: X(\omega) \in B\},~~B \in \bar{\mathcal{B}}$. The distribution function of$\displaystyle X$ is then $\displaystyle \mu(x) = \mathbb{P}\{\omega: X(\omega) \leq x\}$. Now, because for $\displaystyle a < b$, $\displaystyle \mu(b) - \mu(a) = \mathbb{P}\{\omega: a < X(\omega) \leq b\} = \mathbb{P}_X(a,b]$, the distribution $\displaystyle \mu$ corresponds to the measure $\displaystyle \mathbb{P}_X$. We should have $\displaystyle \mu(x) = 1$ as $\displaystyle x \rightarrow \infty$ and $\displaystyle \mu = 0$ as $\displaystyle x \rightarrow -\infty$. Is that of any help?

By the way, which book do you use to study measure theory?

6. If X is a random variable then you can think of $\displaystyle X^{-1}:\mathcal{\overline{B}}\rightarrow \mathcal{F}$. Now $\displaystyle \mu=\mathbb{P}\circ X^{-1}$.

Originally Posted by funnyinga
(1) $\displaystyle 0 \leq \mathbb{P} \leq 1$ for some event A
This is immediately obvious.
(2) $\displaystyle \mathbb{P} (\Sigma) = 1$
We have that $\displaystyle X^{-1}(\overline{\mathbb{R}})=\Omega$, and so this is obvious as well.
(3) For any sequence $\displaystyle A_1, A_2, ...$ which are disjoint
$\displaystyle \mathbb{P} ( \bigcup_i A_i ) = \sum_i \mathbb{P} (A_i).$
Take $\displaystyle A_1,A_2,... \in \overline{\mathcal{B}}$ disjoint, i.e. $\displaystyle \bigcap_{i \in \mathbb{N}}A_i=\emptyset$. This implies that $\displaystyle X^{-1}\left(\bigcap_{i \in \mathbb{N}}A_i\right)=\bigcap_{i \in \mathbb{N}}X^{-1}(A_i)=\emptyset$. As for each i, we have $\displaystyle X^{-1}(A_i) \in \mathcal{F}$ the result follows.