For the probability space $(\Omega, \mathscr F, \mathbb P)$, the random variable $X$ with distribution $F_X$ has Skorokhod representation
$$X(\omega) := \sup \{x \in \mathbb R | F_X(x) < \omega\}$$
It can be shown that in the probability space $(\mathbb R, \mathscr B (\mathbb R), \mathcal L_X)$, the identity random variable
$$Y(\omega) = \omega$$
has distribution $F_Y = F_X$.
Let's take the exponential distribution:
In $((0,1), \mathscr B(0,1), Leb)$, it looks like:
$$X(\omega) = \frac{1}{\lambda}\ln(\frac{1}{1-\omega})$$
or
$$X(1-\omega) = \frac{1}{\lambda}\ln(\frac{1}{\omega})$$
Generally, in $(\Omega, \mathscr F, \mathbb P)$, it looks like:
$$X(\omega) := \sup \{x \in \mathbb R | F_X(x) < \omega\}$$
Specifically, in $(\mathbb R, \mathscr B (\mathbb R), 1-e^{-\lambda x})$, it looks like:
$$X(t) := t$$
The expectations for each are
$$\lambda^{-1} = \int_0^1 \frac{1}{\lambda}\ln(\frac{1}{1-\omega}) d\omega = \int_0^1 \frac{1}{\lambda}\ln(\frac{1}{\omega}) d\omega$$
$$\lambda^{-1} = \int_{\Omega} X d \mathbb P = \int_{\Omega} \sup \{x \in \mathbb R | F_X(x) < \omega\} d \mathbb P(\omega)$$
$$\lambda^{-1} = \int_{\mathbb R} t dF_X(t) = \int_{\mathbb R} t d(1-e^{-\lambda t}) = \int_{\mathbb R} t f_X(t) dt = \int_{\mathbb R} t \lambda e^{-\lambda t} dt$$
So, if $\mu\{x\}=0$, the distribution $\mu$ is absolutely continuous w.r.t $\lambda$.
That is not true. The condition is necessary and sufficient of $X$ to be continuous in the sense that CDF $F$ is a continuous function, but for the existence of a PDF more is needed: $\mu<<\lambda$ which means that $\mu(B)=0$ must be true for every $B\in\mathcal B$ that satisfies $\lambda(B)=0$. This condition is necessary and sufficient for the existence of a Radon-Nikodym derivative. If $f$ is such a derivative (there is more than one) then in your example indeed also the function prescribed by $x\mapsto f(x)$ for $x\in\mathbb R_+$ and $x\mapsto 0$ otherwise is such a derivative, and in that situation functions with that property are commonly chosen as PDF. Next to that we strive to use a PDF having other nice properties like continuity.
I would argue that $\Omega$ can be countable or uncountable, but for a random variable $X$ to be discrete $X(\Omega)$ must be countable - so the random variable has countable support.
For a discrete random variable it is not necessary that $X(\Omega)$ is countable. For example on $(\Omega=\mathbb R,\mathcal B)$ we can define $\mathbb P$ by setting that $\mathbb P(B)=1\iff0\in B$ and then we can define random variable $X$ by $\omega\mapsto\omega$. That leads to discrete random variable with $\mu(\{0\})=1$, but $X(\Omega)=\mathbb R$ is uncountable. In fact if we are looking for a support then $X(\Omega)$ can be a candidate, but is often not okay. It might also be that $X(\Omega)\notin\mathcal B$ and a support (what exactly is it?...) must be something as a "smallest" set $S\in\mathcal B$ with $\mu(S)=1$. By discrete random variables such a set exists and above it is the set $\{0\}$. Actually $X$ is by definition discrete if a countable set $S\subseteq\mathbb R$ exists with $\mathbb P(X\in S)=1$ or equivalently $\mu(S)=1$ and as shown we do not need $X(\Omega)\subseteq S$ for that. If there are elements $s\in S$ with $\mathbb P(X=s)=0$ then we can just "throw them out" and what is left is again a countable subset with $\mu(S)=1$ and now also $\mu(\{s\})>0$ for every $s\in S$, so that a smaller set with $\mu(S)=1$ cannot exist. Since $S$ is countable we have $S\in\mathcal B$ so that $\mu(S)$ is automatically defined.
In the continuous case we must not search for a smallest set $S$ with $\mu(S)$ simply because it does not exist: if $\mu(S)=1$ then also $\mu(S-T)=1$ whenever $T$ is a countable set. We can look for a nice $S\subseteq\mathbb R$ with $S\in\mathcal B$ and $\mu(S)=1$. In the example you mentioned we can take $\mathbb R_+$, but again it is not necessary that $X(\Omega)\subseteq\mathbb R_+$.
Concerning question 2
Let it be that $S\subseteq\mathbb R$ is countable and serves as support of discrete $X$. Then define $\#A$ to be the cardinality of $A\cap S$. Then $\#$ is a measure on $\mathcal B$ (or even on $\wp(\mathbb R)$). By definition:$$\int_A fd\#=\sum_{a\in S\cap A}f(a)$$
Best Answer
Do yo mean to allow equality $X(\omega)=c$, or only the strict inequality $X(\omega)>c$? The argument is the same in either case, but you have two different sets in your question (one allowing $X(\omega)=c$, the other requiring strict inequality). You don't need anything about pushforwards or change of variables. Your composite function $\mathbf{1}_{(c,\infty)}\circ X:\Omega\to\mathbf{R}$ is in fact equal to the characteristic function $\mathbf{1}_{\{\omega\in \Omega:X(\omega)>c\}}$. (Check this!) Since $X$ is a measurable function by definition, the set $\{\omega\in\Omega:X(\omega)>c\}=X^{-1}((c,\infty))$ is measurable, so the characteristic function of this set is measurable and its integral is equal to the measure of the set in question.