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I am studying measure theory by myself, and I often see insane but common definitions such as

A set of random variables $\{X_n\}$ is said to be adapted to a filtration $\{\mathcal{F}_n\}$ if each $X_n$ is measurable with respect to $\mathcal{F}_n$.

But to me, checking $\mathcal{F}$-measurability is exhausting and hard even for a single random variable, let alone countably infinite many of them! How do people in probability deal with this seemingly intractable problem?

For example, even in the trivial cases, such as similar question from Mathstackexchange

Consider the state space $\Omega = \mathbb{R}$, the $\sigma$-algebra, $\mathcal{F} = \{(-\infty, 0], (0, \infty), \varnothing, \mathbb{R}\}$ and the random variable $X : \Omega \rightarrow \mathbb{R}$ defined by \begin{align*} X(\omega) = \begin{cases} 3 & \omega < 0\\ 5 & \omega \geq 0 \end{cases} \end{align*} Is $X$ $\mathcal{F}$-measurable?

Let's think about the computation procedures involved. We need to consider the entire range of $X$, which is $\mathbb{R}$.

Next, we need to recall the definition of measurability,

Given a probability space $(\Omega, \mathcal{F}, P)$, a random variable is a function $X$ from $\Omega$ to the real numbers $\mathbb{R}$ is $\mathcal{F}$-measurable if $\{\omega \in \Omega: X(\omega) \leq x\} \in \mathcal{F}, \forall x \in \mathbb{R}$

Now I have to check for different $x$ in the codomain of $X$. The solution to this problem considered $x < 3$, $3 \leq x < 5$, $5 \leq x$. For each range of values, computed the corresponding inverse image $\{\omega \in \Omega | \cdots\}$, then checked whether if this inverse image is in $\mathcal{F}$.

Ok, so even for this trivial problem, we have at least 1 + 6 steps (one step for dividing up the range, 3 steps for computing each inverse image, 3 steps for checking if each inverse image is in $\mathcal{F}$) we need to do to make sure that $X$ is $\mathcal{F}$ measurable. This obviously lead to exhaustion with even if $X$ was a tiny bit more complicated. Let alone $\infty$ many of them.

So how do people actually show that something is measurable? It couldn't be this complicated could it?

Sin Nombre
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  • The range of $X$ is not all of $\mathbb{R}$, it's just ${3,5}$. – Jürgen Sukumaran Sep 27 '20 at 21:36
  • @TSF Codomain? I don't know the word $X: (\Omega, \mathcal{F}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R})$ – Sin Nombre Sep 27 '20 at 21:38
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    Yes, it's usually called codomain. For you question itself, measurability is something very similar to continuity so you get similar theorems, like composition of measurable functions is measurable. Also, the $\sigma$-algebra $\mathcal{F}$ can be formulated in such a way that all the variables in question are obviously $\mathcal{F}$-measurable or it's not worth it to try, such as with random variables $x$ and $y$ and $\mathcal{F}=\sigma(x)$, the $\sigma$-algebra generated by $x$. – Jürgen Sukumaran Sep 27 '20 at 21:40
  • @TSF Thanks for the pointers. I am interested in big theorems such as $X$ that satisfies property $P$ are all measurable. Or something like $X$ are always measurable with respect to this specific $\mathcal{F}$ – Sin Nombre Sep 27 '20 at 21:41
  • Whether some specific set is measurable can be complicated and difficult. – DanielWainfleet Sep 28 '20 at 01:29
  • The random variable $X$ is always measurable with respect to $\mathcal{F} = \sigma(X)$, the $\sigma$-algebra generated by $X$. We often care more about the Borel $\sigma$-algebra of a given space and then continuity and measurability are really quite similar. – Jürgen Sukumaran Sep 28 '20 at 09:40

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