Probably someone's posted here before on the definition of lim inf, and from comments it appears that that, rather than anything about random variables is the locus of most of what is being asked about. I'll give a definition and some comments, but also throw in some remarks about the application to random variables.
Consider the sequence
$$
4.9,\,6.1,\,4.99,\, 6.01,\, 4.999,\, 6.001,\, 4.9999,\, 6.0001,\,\ldots
$$
The limit inf, or limit inferior, of this sequence is $5,$ and the lim sup is $6.$ This means that for all $\varepsilon>0,$ all but finitely many terms in the sequence are $>5-\varepsilon$ and that is not true of any number bigger than $5,$ and all but finitely many are $<6+\varepsilon$ and that is not true of any number less than $6.$
In probability theory one could say that an "outcome" $\omega$ is randomly chosen and it determines all values $X_1(\omega), X_2(\omega), X_3(\omega),\ldots,$ and their common dependence on the randomly chosen outcome $\omega$ is where they get their randomness.
But for each separate value of $\omega,$ the lim inf and lim sup are just the lim inf and lim sup of a sequence of numbers, defined as above with no reference to randomness.
The dependence of $\liminf\limits_{n\to\infty} X_n(\omega)$ upon $\omega$ makes the lim inf a random variable in its own right, so one might ask what its distribution is, or what its expected value is, an so on.
Here's an exercise (but probably it's more than an exercise): Toss $n$ coins. Forfeit the "tails"; keep the "heads". Repeat until the number of remaining coins is either $0$ or $1.$ Show that the lim inf as $n$ grows, of the probability that it's $0$ rather than $1$ differs from the lim sup by a small positive number (about $10^{-4}$ or $10^{-5},$ I think?).
Best Answer
It's helpful to read the union as "there's some $n$ such that". So the equality you're interested in translates in English to "$\inf X_n < a$ if and only if there's some $n$ such that $X_n < a$."
But that follows easily from the definition of $\inf$. Indeed, suppose $\inf X_n < a$. Since for all $\epsilon > 0$ there's some $n$ such that $X_n - \inf_n X_n < \epsilon$, it follows (with $\epsilon = (a - \inf_n X_n)/2$, for example) that there's some $n$ such that $X_n < a$. Conversely, if there's some $n$ such that $X_n < a$, then clearly $\inf_n X_n < a$ since the $\inf$ is a lower bound.