Let $X$ have the exponential distribution. For $n=1,2,\ldots$ let $Y_{n}$ equal $\lfloor X\rfloor /n$. Prove that $Y_{n}$ is a random variable, and calculate and identify its distribution function.
My approach: To prove that $Y_{n}$ is a random variable it is necessary to verify that $$\{\omega : Y_{n}(\omega)\leq y\}\in \mathscr{F},\ \text{for every}\ y\in\mathbb{R}.$$ Where $\mathscr{F}$ is a $\sigma$-algebra.
Now, $$\{\omega : Y_{n}(\omega)\leq x\}=\{\omega : \frac{\lfloor X(\omega)\rfloor}{n}\leq y\}=\bigcup_{i=1}^{n}\{\omega: \lfloor X(\omega)\rfloor\leq yi\}$$ Since $X(\omega)$ is a random variable and $iy\in\mathbb{R}$, for each $i=1,2,\ldots$, then every element of the precedieng union is an element of the $\sigma$-lagebra, hence $$\bigcup_{i=1}^{n}\{\omega: \lfloor X(\omega)\rfloor\leq yi\}\in \mathscr{F}.$$
Is this a good reasoning?.
As for the distribution: I know that $Y_{n}$ is a descrete R.V. and since
$l\leq X\leq l+1$, then
\begin{equation}F(x)=P(\omega :X(\omega)\leq yn +1)=P(\omega :X(\omega)< [yn] +1)\end{equation}
I know this distribution is a geometric one, but how can I find it.
