So assume we have a probability space $(\Omega, \mathcal{F}, P)$ and a random variable $X : \Omega \rightarrow \mathbb{R}^*$. We can derive from this a distribution $P_X$, and a distribution function $F$. My book says that when working with distribution function directly we have no need to know the underlying probability space since everything is well determined. We just need to show that we can find one. So here is my attempt :

  • Here we start with the distribution function $F$. increasing and right continuous, st. $F(\infty) = 1$ and $F(-\infty) = 0$. We define :

\begin{align} P_I : I &\longmapsto [0,1] \quad I \in \mathbb{R}^2 ( interval) \\ [a,b] &\longmapsto F(b) - F(a) \end{align}

  • We can define $P_X$ on the borel sets by decomposing the set into disjoint intervals :

\begin{align} P_X : \mathcal{B}(\mathbb{R}^*) &\longmapsto [0,1] \quad B = \dot \bigcup B_i \quad B_i \in \mathbb{R}^2 \\ B &\longmapsto \sum_i P_I[a_i,b_i] \end{align}

  • By taking $X$ as the identity function we have $\Omega = \mathbb{R}^*$ and $P = (P_X)_{\Omega}$

I suspect that I can use the fact that F is increasing and right continuous to say that we can find a measure $\mu$ s.t $\mu(a,b) = F(b)-F(a)$ and then use this measure as probability measure. But I'm scared to miss any vital step along this way.

I would like to know if there is any flaw in my reasoning. Or anything missing. Thanks for any input !

  • $\begingroup$ Every Borel set is not a countable union of disjoint closed intervals. $\endgroup$ – Did Aug 7 '14 at 20:46
  • $\begingroup$ How so ? The sigma algebra generated by the open intervals of $\mathbb{R}$ form the borel sigma algebra of $\mathbb{R}$. It is closed under union and one can always form a disjoint union of sets starting from a union of sets. Am I missing something ? $\endgroup$ – user149705 Aug 7 '14 at 20:57
  • $\begingroup$ The Borel sigma-algebra is generated by the open intervals. Countable unions of intervals are Borel sets. But every Borel set is not a countable union of intervals. $\endgroup$ – Did Aug 7 '14 at 21:01
  • $\begingroup$ Oh ok ! How would you solve this to my specific problem ? $\endgroup$ – user149705 Aug 7 '14 at 21:21
  • $\begingroup$ "I would like to know if there is any flaw in my reasoning." Done. If you have another question, you may want to write another post (or to grab any measure theory textbook, since they all explain this construction). $\endgroup$ – Did Aug 7 '14 at 21:23

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