# What does it mean for a sequence of random variables to be increasing?

According to a textbook I'm reading, one statement of the monotone convergence theorem is:

If $X_1, X_2, ...$ is a sequence of non-negative random variables increasing to the random variable $X$, then the expectations $E[X_1], E[X_2], ...$ increase to $E[X]$.

I'm having trouble understanding exactly what this means. How can a sequence of random variables be "increasing"? What does it mean for a random variable to be "greater than" another random variable? I thought that random variables were in essence just functions. I guess you could say a function $f(x)$ is greater than another function $g(x)$ if $\forall x \in \mathbb{R}, f(x) \ge g(x)$. But I'm getting confused when thinking about it in context of random variables.

I apologize if this is self-evident and I'm just misunderstanding what the theorem is saying... Thanks in advance!

• Your thought regarding random variables as functions is exactly correct. What this means is that for every outcome $O$ in the sample space, we have $$X_1(O) \le X_2(O) \le X_3(O) \le ...$$ Feb 16, 2016 at 21:12
• @LeeMosher What exactly does $X_1(O)$ mean for a random variable $X_1$? Feb 16, 2016 at 21:16
• It's what you said: $X_1$ is a function. The domain of this function is the sample space. So, input an element of the sample space, call that input $O$. The random variable outputs a number, denoted $X_1(O)$, as is usual for functions. Feb 16, 2016 at 21:17
• To put it in concise notation, $X_1$ is a function with domain $S =$ the sample space, and with range $\mathbb{R} =$ the real numbers, that is, $X_1 : S \to \mathbb{R}$. Feb 16, 2016 at 21:18

Recall that a random variable can be defined as a measurable function $X:\Omega \to \mathbb{R}$, so in this case the definition that you gave in your question works perfectly well, expressed as $$X_n(\omega)\leq X_{n+1}(\omega)$$ for all $n\geq1$ and all $\omega \in\Omega$.
In the case of the Monotone Convergence Theorem though, this condition is actually stronger than necessary. The sequence of random variables $(X_n)_{n\geq 0}$ need only be increasing almost surely, that is, $$\mathbb{P}(\{\omega \in \Omega:X_n(\omega)\leq X_{n+1}(\omega) \text{ for all n}\})=1.$$