# On Definition of Random Variables

Given a probability space $$(\Omega, \mathcal{F}, P)$$, a random variable X is defined to be a function from $$\Omega$$ to $$\mathbb{R}$$ such that for every $$a \in \mathbb{R}$$, the set $$\{X \leq a\} := \{ω \in \Omega : X(\omega) ≤ a\}$$ belongs to $$\mathcal{F}$$.

From the axiomatic definition of a probability space, for every $$a \in \mathbb{R}$$, if the set $$\{X \leq a\}$$ belongs to $$\mathcal{F}$$, the set $$\{X > a\}$$ must also belong to $$\mathcal{F}$$. Consequently, the set $$\{a < X \leq b\}$$ must also belong to $$\mathcal{F}$$.

I want to show that the set $$\{a \leq X < b\}$$ also belongs to $$\mathcal{F}$$ as well, so I can generalize it to any other set, but I have not managed to prove it.

Anyone has an idea on how to do it?

• This might be a way to think about it, although not necessarily the simplest way: we can find an infinite sequence $\{a_i\}$ that converges to $a$ from below. Therefore, $\bigcup_{i=1}^\infty \{X\le a_i\}=\{X<a\}$. This set is also in $\mathcal{F}$, because $\mathcal{F}$ is $\sigma$-algebra. Jun 15 at 1:08

## 3 Answers

As @whoknowsnot has shown, for every $$a \in \mathbb R$$, the event $$\{X< a\}\in \mathcal F$$, and since $$\{X\leq a\}\in \mathcal F$$ by definition, the difference of these two events, namely $$\{X = a\}$$, is also in $$\mathcal F$$. Hence, from the result that you have already proven, namely that $$\{a < X \leq b\} \in \mathcal F$$, it is easy to deduce that \begin{align} \{a < X < b\} &\in \mathcal F,\\ \{a \leq X < b\} &\in \mathcal F,\\ \{a \leq X \leq b\} &\in \mathcal F, \end{align} just by adding in or taking away the singleton events $$\{X = a\}$$ and $$\{X = b\}$$.

To quote from Wikipedia, a $$\sigma$$-algebra is defined as

1. $$\Omega$$ is in $$\mathcal F$$, and $$\Omega$$ is considered to be the universal set in the following context.
2. $$\mathcal F$$ is closed under complementation: If $$A$$ is in $$\mathcal F$$, then so is its complement, $$A^c$$.
3. $$\mathcal F$$ is ''closed under countable unions'': If $$A_1$$, $$A_2,\ldots$$ are in $$\mathcal F$$, then so is $$A=A_1 ∪ A_2 ∪ \cdots$$

From these properties, it follows that the σ-algebra is also closed under countable intersection (by applying De Morgan's laws).

You can write the desired set using only negation and countable unions of the foundational sets as:

\begin{align} \{ a \leq X < b \} &= \{ a \leq X \leq b \} - \overline{ \{ X \leq b \}} \\[6pt] &= \{ X \leq b \} - \bigcup_{n=1}^\infty \bigg\{ X \leq a - \frac{1}{n} \bigg\} - \overline{ \{ X \leq b \}}. \\[6pt] \end{align}

Since the sigma-field $$\mathcal{F}$$ is closed under negation, countable unions and set difference (based on the foundation sets) this is sufficient to prove that $$\{ a \leq X < b \} \in \mathcal{F}$$.