What is the reasoning behind this step in the proof of $\lim_{x \to a^- }F_X(x)= P(X\lt a)$

I have two questions about the reasoning behind the steps in the following proof:

Suppose $$F$$ is the cumulative distribution function of a random variable $$X.$$ Then $$\lim_{x\to a^-}F_X(x)=P(X \lt x)$$

Proof:(written a bit informally)

Suppose $$F_X$$ is the CDF of a random variable $$X$$

$$F_X$$ being a CDF $$\implies$$

$$\implies F_X \text{ is monotonic increasing} \tag 1$$

$$\implies \lim_{x\to a^-}F_X(x) = \lim_{n \to \infty}F_X \left(a- \frac{1}{n}\right) \tag 2$$

By definition $$F_X(a-\frac{1}{n})=P(E(X\leq a-\frac{1}{n})) \tag 3$$

Notice that $$E_n= E(X\leq a-\frac{1}{n})$$ is an increasing sequence of events so by the lemma

Lemma: If $$E_N$$ for $$n\geq 1$$ is an increasing sequence of events in a probability space and $$E=\bigcup_{i=1}^\infty E_i$$, Then $$P(E)=\lim_{n\to \infty}P(E_N)$$.

therefore by the lemma:

$$\lim_{x \to a^-}F_X(x)= \lim_{n\to \infty}P\left( E\left(X\leq a-\frac{1}{n}\right)\right) = P\left( \bigcup_{i=1}^\infty E\left(X\leq a -\frac{1}{i} \right) \right).$$

However,

\begin{align} & \bigcup_{i=1}^\infty E(X\leq a -\frac{1}{i}) \tag 4 \\[8pt] = {} & \{\omega \in \Omega: X(\omega) \leq a -\frac{1}{i} \space \text{for some } i \geq 1\} \tag 5 \\[8pt] = {} & \{\omega \in \Omega: X(\omega) < a\} \tag 6 \\[8pt] = {} & E(X < a) \end{align}

Thus, $$\lim_{x\to a^-}F_X(x)=P(E(X < a))=P(X < a)$$

$$\square$$

My questions are:

1) How does step [1] imply step [2] exactly?

2) Why is is that $$\bigcup_{i=1}^{\infty}E(X\leq a -\frac{1}{i})$$

$$=\{\omega \in \Omega: X(w) \leq a -\frac{1}{i} \space \text{for some } i \geq n\}$$

3) How does step [4] imply step [5]??

• I have no idea why you have expectations in this proof. The proof seems overcomplicated as well. Commented Dec 30, 2019 at 19:51
• At the beginning, I think you'd want to use specific continuity properties of the CDF (besides just monotone increasing, e.g., right continuous). That property motivates the $x \to a^{-}$ limit Commented Dec 30, 2019 at 19:53
• @EpsilonDelta $E(X \in A)$ is used to mean $\{ \omega \in \Omega: X(\omega) \in A \}$ where A is some subset of $\mathbb{R}$ Commented Dec 31, 2019 at 6:42

The way in which $$(1)$$ implies $$(2)$$ is by implying that the first limit in $$(2)$$ exists. Once you have that, the equality in $$(2)$$ must hold.
Your way of using the capital $$E$$ is new to me, but if I understand it correctly, the implication from $$(4)$$ to $$(5)$$ is merely the definition of "union", provided what you meant in $$(5)$$ is "for some $$i\in\{1,2,3,\ldots\}.$$" And if you didn't mean that, then there's the question of what the "$$n$$" is, in step $$(5),$$ and that is not at all clear. To say that a point is in the union of a certain collection of sets means just that it is a member of at least one of those sets.
Let $$X$$ be a real-valued random variable with CDF $$F_X$$ and $$a\in\mathbb{R}$$. We have $$\bigcup_{n\ge0}\left\{\omega\in\Omega\colon X(\omega)\le a-\frac{1}{n}\right\}=\{\omega\in\Omega\colon X(\omega) If $$\omega\in\Omega$$ is such that $$X(\omega)\le a-\frac{1}{n}$$, then $$X(\omega)\le a-\frac{1}{n}, so the inclusion "$$\subseteq$$" is clear. If, on the other hand, $$\omega\in\Omega$$ is such that $$X(\omega), note that $$X(\omega) for $$n=\left\lceil\frac{2}{a-X(\omega)}\right\rceil$$, so the reverse inclusion holds too. Alternatively, one could argue that if $$X(\omega)>a-\frac{1}{n}$$ for all $$n\ge0$$, then, by taking the limit as $$n\rightarrow\infty$$, $$X(\omega)\ge a$$, which also proves the reverse inclusion by contraposition. Now, since $$F_X$$ is monotone, the limit $$\lim_{x\rightarrow a^-}F_X(x)$$ exists. Therefore, and by invoking the lemma, we have $$\lim_{x\rightarrow a^-}F_X(x)=\lim_{n\rightarrow\infty}F_X\left(a-\frac{1}{n}\right)=\lim_{n\rightarrow\infty}\mathbb{P}\left(X\le a-\frac{1}{n}\right)=\mathbb{P}(X