probability theory - The cdf and pdf of the random variable $X(omega)=1/omega$

Consider the probability space $(\Omega,\mathcal{F},\mathbb{P})$ where $\Omega=(0,1]$, $\mathcal{F}$ is the Borel $\sigma$-field generated by intervals of the form $(0,\frac{b}{2^n}]$ with $b\leq 2^n$, $b\in\mathbb{N}$, and $\mathbb{P}$ is the uniform Lebesgue measure. We define the real-valued random variable $X(\omega)=\frac{1}{\omega}$.

I'm struggling a little bit to derive the cumulative distribution function and probability density function of $X$.

My attempt: $F(x)=\mathbb{P}(\omega\in\Omega:X(\omega)\leq x)=\mathbb{P}(\omega\in\Omega:\frac{1}{\omega}\leq x)=\mathbb{P}(\omega\in\Omega:\omega\geq \frac{1}{x})$ for $x\in\mathbb{R}_{\geq 1}$.

In the case $x<1$, we get $\mathbb{P}(\emptyset)=0$. That's because for small values of $x$, $1/x$ explodes but $\omega$ can take values up to $1$.

So, $F(x)=\frac{1}{x}\mathbb{I}_{x\geq 1}$ where $\mathbb{I}$ is the indicator function.

Then the probability density function is given by

$$f_X(x)=\frac{d}{dx}F_X(x)=-\frac{1}{x^2}\mathbb{I}_{x\geq 1}.$$

Is my reasoning correct? I'm not sure how $\mathcal{F}$ plays any role here. I'd appreciate any hints.

Answer

If $x\ge 1$ then

$$\begin{align} \Pr(X\le x) & = \Pr\left( \left\{\omega : \omega\ge \frac 1 x \right\} \right) = 1 - \frac 1 x, \\[10pt] \text{So } \frac d {dx} \Pr(X\le x) & = \frac d {dx} \, \left( 1 - \frac 1 x \right) = \frac 1 {x^2}. \end{align}$$ You have $1/x$ where you needed $1 - (1/x).$ Note that a probability density function cannot be negative, as your proposed density function is.