The problem I have been presented with is as follows:
Let $\Omega = (0,1)$, $\mathcal{F} = \mathcal{B}(\Omega) $, $P = \lambda \text{ (Lebesgue measure)}$. Assume $X(\omega) = \omega^{-3}$. Prove $X$ is a random variable, and find the CDF and PDF for $X$.
Here is my attempt:
Let $A = \{ \omega \in (0,1) \text{ } \vert \text{ } a < X(\omega) < b\}$ where $0 < a < b$. We may rewrite $A$ as $$ A = \{ \omega \in (0,1) \text{ } \vert \text{ } a < \omega^{-3} < b\} $$ which is to say $$A = \{ \omega \in (0,1) \text{ } \vert \text{ } b^{-\frac{1}{3}} < \omega < a^{-\frac{1}{3}} \}$$ which is a Borel set in $\mathcal{F}$. Thus $P(A) = a^{-\frac{1}{3}} - b^{-\frac{1}{3}}$. Therefore, $X$ is a random variable (I understand this is quite hand-wavy and leaves a lot to be desired, but this is the final line in similar "proofs" from our lecture).
This is where I struggle. Is the CDF for $X$ just simply $F_X(u) = u^{-\frac{1}{3}}$, giving the PDF for $X$ as $f_X(u) = \frac{1}{3} u^{-\frac{4}{3}}$? I thought this was the solution until I realized $\lim_{u \to \infty} F_X(u) \neq 1$, which should equal $1$ for any CDF by definition. Where is my understanding failing?
After reading the provided hints, I think I have created a full answer:
First: Let $A = \{ w \in \Omega = (0,1) \text{ } \vert \text{ } a < X(\omega) < b \}$ where $0 < a < b$. From $a < X(\omega) = \omega^{-3}$, we have $w < a^{-\frac{1}{3}}$. Similarly, from $X(\omega) = \omega^{-3} < b$, we have $b^{-\frac{1}{3}} < \omega$. Thus, we have that $$ A = \{ w \in \Omega = (0,1) \text{ } \vert \text{ } b^{-\frac{1}{3}} < \omega < a^{-\frac{1}{3}} \} = (b^{-\frac{1}{3}},a^{-\frac{1}{3}}) $$ Hence, $A$ is an interval, and any interval is a Borel set in $\mathcal{F}$. Therefore, $X$ is a measurable function.
Second: For $u \geq 1$, we have \begin{align} F_X(u) &= P(X \leq u) \nonumber\\ &= P( \{ w \in (0,1) \text{ } \vert \text{ } \omega^{-3} < u \} ) \nonumber\\ &= P( \{ w \in (0,1) \text{ } \vert \text{ } u^{-\frac{1}{3}} \leq \omega \} ) \nonumber\\ &= 1 - P( \{ w \in (0,1) \text{ } \vert \text{ } \omega \leq u^{-\frac{1}{3}} \} ) \nonumber\\ &= 1 - u^{-\frac{1}{3}}\nonumber \end{align} Thus, the CDF for $X$ is given as $$ F_X(u) = \begin{cases} 0 & \text{ for } u < 1 \\ 1 - u^{-\frac{1}{3}} & \text{ for } u \geq 1 \\ \end{cases} $$ From this expression, we may find the PDF of $X$ as
$$ f_X(u) = \begin{cases} 0 & \text{ for } u < 1 \\ \frac{1}{3} u^{-\frac{4}{3}} & \text{ for } u \geq 1 \\ \end{cases} $$