Replacing continuous variables in a limit with a sequence

limitsmeasure-theoryprobability distributionsprobability theoryprobability-limit-theorems

I have a question regarding the nuts and bolts involved in the proof of the limit of CDFs. The statement is that

Proposition: Let $X$ be a random variable with CDF $F_X(.)$. Then $F_X(.)$ posses the following property.

$$\lim_{x \to \infty} F_X(x) = 1$$

Proof:

Consider a sequence $\{x_n\}$ with $ n \in \mathbb{N}$ such that it monotonically increases to $\infty$. Then we have

\begin{eqnarray}
\lim_{x \to \infty}F_X(x) &=& \lim_{x \to \infty} \mathbb{P}(X \leq x) \\
&=& \lim_{n \to \infty} \mathbb{P}(X \leq x_n) \label{eqnref} \\
&=& \mathbb{P} \left \{ \bigcup_{n \in \mathbb{N}} \{ω : X(ω) ≤ x_n \} \right \} \\
&=& \mathbb{P}(\Omega) \\
&=& 1.
\end{eqnarray}

My question is regarding the second step where the continuous variable $x$ is replaced by the member of a sequence $x_n$. I feel lack of rigor in this step. To be precise, my questions are

  1. Why is this step valid?
  2. The trajectory that $x$ can take while approaching $\infty$ are many, while the sequence $x_n$ is assumed to be monotonically increasing. How do we know for sure that this difference in the way to approach infinity will not change the limit?
  3. Is there a way to make the proof look more rigorous as in is there a rigorous way to substantiate this step of replacing $x$ with $x_n$?

Please help.

Best Answer

  1. This step is valid as long as proving it is obvious/easy/possible. This could be discussed as what is obvious of experimented people may be a full-fledge exercice for beginners, but in any case, it is true.

  2. I understand your point of multiple trajectories for $x$, but take into account that for any of these multiple trajectories, you can extract a monotonically increasing one.

  3. Using the definition of these limits could help to clarify :

$$ \lim_{x \to \infty} \mathbb{P}(X \leq x) = l \iff \forall \epsilon >0, \exists A \mid x > A \Rightarrow | \mathbb{P}(X \leq x) - l | < \epsilon$$

$$ \lim_{n \to \infty} \mathbb{P}(X \leq x_n) = l \iff \forall \epsilon >0, \exists N \mid n > N \Rightarrow |\mathbb{P}(X \leq x_n) - l | < \epsilon $$

So, what you need is to find a way from a $A$ (resp $N$) large enough to a have the nice property, to find a $N$ (resp $A$) large enough to have the other nice property. In order to do so, I would write the definition of : $ \lim_{n \to \infty} x_n = +\infty$.

Related Question