Proving formally $\lim_{x \to -\infty}\mathrm{Pr}( \left \lfloor{x}\right \rfloor \le X < x) = 0$ (Proof check)

Question

we have

$$\lim_{x \to -\infty}\mathrm{Pr}( \left \lfloor{x}\right \rfloor \le X < x) $$ where X is a real random variable, and $x \in R$.

My idea of a proof would be by contradiction:

Assume $\exists$ an $\epsilon_0 $ for witch $|\mathrm{Pr}( \left \lfloor{x}\right \rfloor \le X < x)|$ is not < $\epsilon_0 $ for any $x < M$.

Then set $\mathrm{Pr}( \left \lfloor{x}\right \rfloor \le X < x) = c > \epsilon_0 $ but there are countably many $\mathrm{Pr}( \left \lfloor{x}\right \rfloor \le X < x)$ that must all be $\ge c$.

This lead to the desired contradiction because eventually the sum of countably many increasing numbers will be > 1.

This proof I wrote seems pretty raw (I am even unsure it is correct), what could I do to improve it?

are there other faster proofs? maybe not by contradiction?

Thanks in advance

Answer 1

First prove formally that $\lim_{x\rightarrow-\infty}P\left(X<x\right)=0$. (Do you know how?)

Let $\epsilon >0$.

Find some $x_0$ such that $Pr(X<x_0)<\epsilon$.

Then for $x\leq x_0$ we have $$Pr(\lfloor x\rfloor\leq X<x)\leq Pr(X<x_0)<\epsilon$$

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edit:

From $P\left(X<x\right)\leq P\left(X\leq\lceil x\rceil\right)$ it follows that it is enough to prove that $\lim_{n\rightarrow\infty}P\left(X\leq-n\right)=0$ where $n$ only take values in $\mathbb{N}=\left\{ 0,1,2,\dots\right\} $. It is obvious that:

$$P\left(X\leq0\right)=\sum_{k=1}^{n}P\left(-k<X\leq-k+1\right)+P\left(X\leq-n\right)$$

If $n\rightarrow\infty$ then: $$\sum_{k=1}^{n}P\left(-k<X\leq-k+1\right)\rightarrow\sum_{k=1}^{\infty}P\left(-k<X\leq-k+1\right)=P\left(X\leq0\right)$$ Consequently: $$P\left(X\leq-n\right)\rightarrow0$$

Answer 2

I suggest proceeding as follows:
Let $P_A(x) := P(\lfloor x \rfloor \le X < x)$. If $P_A(x_n) > \epsilon \ \forall n \in\mathbb N$ and some sequence $x_n\to-\infty$ we find that $$1 = P(X\in \mathbb R) \ge \sum_{n=1}^{\infty} P_A(x_n) > \sum_{n=1}^\infty \epsilon = \infty$$ wich is a contradiction. Thus for any sequence $x_n\to-\infty$, $P_A(x_n) \to 0$ necessarily. This proves $$\lim_{x\to-\infty} P_A(x) = 0$$ as claimed.