Interpreting probability question in terms of measure theory 
Let $X$ and $Y$ be real valued random variables with joint pdf$$f_{X,Y}(x,y)=\begin{cases}\frac{1}{4}(x+y), & 0\leq x\leq y\leq 2 \\ 0, & \text{ otherwise }\end{cases}
$$
Calculate the probability $\mathbb{P}\{Y<2X\}$.

I am trying to view this problem in a measure-theoretic perspective and I am wondering if I am thinking about this properly.
Let $(\Omega ,\mathcal{F},\mathbb{P})$ be a probability space on which we define two random variables (measurable functions) $X$ and $Y$. We then consider their joint distribution, the pushforward measure of the random variable $T(\omega )=(X(\omega ),Y(\omega ))$ on $\left (\mathbb{R}^2,\mathcal{B}\times \mathcal{B}\right )$ defined by$$T_\star \mathbb{P}(A)=\mathbb{P}\left (T^{-1}(A)\right )=\mathbb{P}((X(\omega ),Y(\omega ))\in A).$$Then by the question, we have that $\dfrac{dT_\star \mathbb{P}}{d\lambda}=f_{X,Y}$ i.e. the Radon-Nikodym derivative of the pushforward of $T$ with respect to the Lebesgue measure is the pdf of $(X,Y)$.
How can I then calculate $\mathbb{P}\{Y<2X\}$? Somehow I have to relate the probability of this set to the pushforward measuere of which I know the density. What is the theorem that allows me to relate these two measures?
Essentially, im looking for the measure of the set $\{\omega :Y(\omega )<2 X(\omega )\}\subset \Omega$. I know the density of a measure on $\mathbb{R}^2$ which is a different set than $\Omega$. How do I know that $\{(x,y):y<2x\}\subset \mathbb{R}^2$ is the subset of $\mathbb{R}^2$ with which I need to integrate over?
 A: 
How do I know that $\{(x,y): y < 2x \} \subset \mathbb{R}^2$ is the subset of $\mathbb{R}^2$ with which I need to integrate over?

This follows from the definition of the term "density" under the measure-theoretic framework.  See, for example, Probability and Measure (Section 20, equation (20.9)), which I quote as follows:

A random variable and its distribution have density $f$ with respect to Lebesgue measure if $f$ is a nonnegative Borel function on $\mathbb{R}^1$ and
\begin{equation*}
P[X \in A] = \mu(A) = \int_A f(x) dx, \quad A \in \mathscr{R}^1.
\end{equation*}

Although this (fundamental) definition applies to a single random variable, it clearly can be generalized to random vectors, as the author later sketched in the same section:

The distribution (of a random vector) may as for the line be discrete in the sense of having countable support.  It may have density $f$ with respect to $k$-dimensional Lebesgue measure: $\mu(A) = \int_A f(x) dx$. As in the case $k = 1$, the distribution $\mu$ is more fundamental than the distribution function $F$, and usually $\mu$ is described not by $F$ but by a density or by discrete probabilities.

In your case, your $A$ is clearly the two-dimensional Borel set $\{(x, y): y < 2x \}$, whence, according to the definition above,
\begin{align*}
P[(X, Y) \in A] = \mu(A) = \int_A f(x, y) dxdy. 
\end{align*}
A: In probability space $(\Omega, \mathcal{F}, \mathbb{P})$ with variables $X,Y$ with joint distribution measure $\mu_{X,Y}$:
\begin{align*}
 P \{ Y < 2X \} &= \int_\Omega \mathbb{I}_{y < 2x} \, d\mathbb{P} \\
 &= \int_{\mathbb{R}^2} \mathbb{I}_{y < 2x} \, d\mu_{X,Y} \\
\end{align*}
The definition of a joint density function is the nonnegative Borel measurable $f_{X,Y}(x,y)$ such that:
\begin{align*}
  \mu_{X,Y}(C) = \int_{-\infty}^\infty \int_{-\infty}^\infty \mathbb{I}_C(x,y) f_{X,Y}(x,y) \, dy \, dx \quad C \in \mathcal{B}(\mathbb{R}^2) \\
\end{align*}
Then $d\mu_{X,Y} = f_{X,Y} \, dx \, dy$, so plugging that in:
\begin{align*}
 P \{ Y < 2X \} &= \int_{\mathbb{R}^2} \mathbb{I}_{y < 2x} \, f_{X,Y} \, dx \, dy \\
&= \int_{\mathbb{R}^2} \mathbb{I}_{y < 2x} \, \mathbb{I}_{0 \le x \le y \le 2} \frac{1}{4}(x+y) \, dx \, dy \\
&= \int_0^2 \int_{y/2}^y \frac{1}{4}(x+y) \, dx \, dy \\
&= \int_0^2 \frac{7}{32} y^2 \, dy \\
&= \frac{7}{12} \\
\end{align*}
