Do we explicitly need to assume that a cumulative distribution function satisfies $F(x)\xrightarrow{x\to\infty}1$?

Question

Remember that $F:\mathbb R\to[0,1]$ is called distribution function if

1. $F$ is nondecreasing;
2. $F$ is right-continuous;
3. $F(x)\xrightarrow{x\to-\infty}0$;
4. $F(x)\xrightarrow{x\to\infty}1$.

Now, drop condition (4.) for a moment. Let $$F^{-1}(u):=\inf\{x\in\mathbb R:F(x)\ge u\}\;\;\;\text{for }u\in(0,1).$$ We can show that $$F^{-1}(u)\le x\Leftrightarrow u\le F(x)\;\;\;\text{for all }x\in\mathbb R\text{ and }u\in(0,1)\tag1$$ and hence $$\{u\in(0,1):F^{-1}(u)\le x\}=(0,F(x)]\cap(0,1)\;\;\;\text{for all }x\in\mathbb R.\tag2.$$ Using this we should immediately be able to conclude that $$X:=F^{-1}$$ is a random variable on $((0,1),\mathcal B((0,1)),\mathcal U_{(0,\:1)})$, where $\mathcal U_{(0,\:1)}$ denotes the uniform distribution on $(0,1)$, with $$\operatorname P\left[X\le x\right]=F(x)\;\;\;\text{for all }x\in\mathbb R\tag3,$$ which is to say that $F$ is the distribution function of $X$.

However, since any measure is continuous from below and $$\mathbb R=\bigcup_{x\in\mathbb R}(-\infty,x]\tag4,$$ the left-hand side of $(3)$ needs to approach $1$ as $x\to\infty$ and hence condition (4.) is satisfied.

So, am I missing something or is (4.) an immediate consequence of (1.)-(3.)?

Answer 1

If $F \colon \mathbb R \to [0,1]$ is nondecreasing, then as $x \to \infty$, $F(x)$ should approach some limit $L$ even without condition 4: that limit $L$ is the least upper bound on $F$. Condition 4 just tells us that $L=1$.

But if we don't have condition 4, then $$F^{-1}(u):=\inf\{x\in\mathbb R:F(x)\ge u\}$$ does not give a real number for all $u \in (0,1)$: only for $u \in (0,L)$ and possibly for $u=L$. For $u \in (L,1)$, $F^{-1}(u) = +\infty$.

I can imagine it sometimes being reasonable to have a quasi-CDF without condition 4 represent a $(\mathbb R \cup \{+\infty\})$-valued random variable $X$, with the limit $L$ giving us $\Pr[X \text{ is finite}]$. But you're not going to get a real-valued random variable this way.

To see examples of this happening, take any CDF and divide it by $2$ to get a quasi-CDF that does not satisfy condition 4, and approaches $\frac12$ in the limit as $x \to \infty$.