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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?

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    $\begingroup$ 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. $\endgroup$ Jun 15 at 1:08

3 Answers 3

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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 $\{X = b\}$.

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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).

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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 $\{ a \leq X < b \} \in \mathcal{F}$.

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