Distribution function of $ X(\omega):=\frac{1}{\lambda} \ln \frac{1}{1-\omega},$ in Lebesgue measure I am having trouble in finding the right approach to this exercise:

Define the probability space $(\Omega, \mathcal{A},P) =((0,1), \mathcal{B}, \mu)$ with $\mu$ as the Lebesgue measure and $ \mathcal{B}$ the Borel-$\sigma$-algebra. 
  Find the distribution function of the random variable $$ X(\omega):=\frac{1}{\lambda} \ln \frac{1}{1-\omega},$$
  where $\lambda$ is a positive parameter.

I know that the definition of a distribution function is $P(X\leq x) $, but to be honest I'm a bit overwhelmed and don't know how to start. Does it have to do with Lebesgue integration? Because that's what we were doing last week in the lecture. Thanks in advance.
 A: Let's take this one step at a time.
You have to figure out $P(X\leq x)$, the probability of the set $\{\omega\mid X(\omega)\leq x\}$, right? Well, let's try and get a feel for what set that is, first of all. It's going to be a subset of $(0, 1)$, so we're just talking about a set of real numbers here, maybe an interval or something.
So, for which $\omega$ do we have $\frac{1}{\lambda} ln \frac{1}{1-\omega}\leq x$, where $x$ is a constant in $(0, 1)$? This is just solving an inequality. At the end of it, you end up with some set $\{\omega\mid X(\omega)\leq x\}$.
Now, what is the probability of that set? Well, the probability measure in play here is just the Lebesgue measure $\mu$. You're being asked to compute $\mu(\{\omega\mid X(\omega)\leq x\})$, in other words, just the length of that set.
A: Definition $$P(X \le x) := P(\omega | X(\omega) \le x)$$
Two steps in computing $P(X \le x)$:


*

*Find all $\omega$ s.t. $X(\omega) \le x$

*Compute the probability of all those $\omega$'s.
For $x \le 0$, $P(X\leq x) = P(\emptyset) = 0$
Let $\omega \in \Omega = (0,1)$. For $x > 0$, $X(\omega) \le x$
Step 1:
$$ \iff \frac1{\lambda}\ln(\frac{1}{1-\omega}) \le x$$
$$ \iff \omega \le \frac{e^{\lambda x} - 1}{e^{\lambda x}}$$
$$ \iff \omega \in (0,1) \cap (-\infty,\frac{e^{\lambda x} - 1}{e^{\lambda x}})$$
$$ \iff \omega \in (0,\frac{e^{\lambda x} - 1}{e^{\lambda x}})$$
Step 2:
$$\mu(\omega | \omega \in (0,\frac{e^{\lambda x} - 1}{e^{\lambda x}}))$$
$$= \mu((0,\frac{e^{\lambda x} - 1}{e^{\lambda x}}))$$
$$= \frac{e^{\lambda x} - 1}{e^{\lambda x}}$$
Therefore, $F_X(x) = P(X \le x) = (1-e^{-\lambda x})1_{x \ge 0} \to f_X(x) = \lambda e^{-\lambda x}1_{x \ge 0}$, which we recognise as the exponential distribution.
