Marginal Distributions from Joint Distribution Here's a seemingly common proof for the formula of a marginal distribution using a bivariate joint distribution, for which I'm not clear on each step:
Setup: Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $X, Y$ be jointly continuous random variables.  Then for any Borel subset $B$ of $\mathbb{R}^2$,
$$
P((X,Y) \in B) = \int_B f_{X,Y} \, d\lambda^2,
$$
where $\lambda^2$ is the Lebesgue measure on $\mathbb{R}^2$ and $f_{X,Y} : \mathbb{R}^2 \to \mathbb{R}$ is the joint probability density of $(X,Y)$.  
Claim: (Consider $X$): $X$ is continuous with probability density function
$$
f_X(x) = \int_{\mathbb{R}} f_{X,Y}(x,y)\, d\lambda(y) \qquad \text{for all } x \in \mathbb{R}.
$$
(Typical) Proof: Consider $X$ and let $A$ be a Borel set in $\mathbb{R}$.  Then
\begin{align}
P(X \in A) & = P(X \in A, Y \in \mathbb{R}) &\qquad (1)\\
& = P((X,Y) \in A \times \mathbb{R}) &\qquad (2) \\
& = \int_{A \times \mathbb{R}} f_{X,Y}(x,y) d\lambda^2 &\qquad (3) \\
& = \int_A \int_{\mathbb{R}} f_{X,Y}(x,y) d\lambda(y) d\lambda(x) &\qquad (4) \\
& = \int_A f_X(x) d\lambda(x). &\qquad (5)
\end{align}
My attempt at explaining each line:
(1) The events $\{X \in A\}$ and $\{Y \in \mathbb{R}\}$ are independent and $P(Y \in \mathbb{R}) = 1$, so $P(X \in A, Y \in \mathbb{R}) = P(X \in A)P(Y \in \mathbb{R}) = P(X \in A).$
(2) Definition of joint probability distribution measure
(3) Definition of joint probability density
(4) Tonelli's theorem
(5) Not sure here.  Somehow $f_X(x) = \int_{\mathbb{R}} f_{X,Y}(x,y) \, \mathrm{d} \lambda(y)$.  Certainly this is a function of $x$, but why must it be the probability density of $X$?
Do these seem correct?

Update:  Based on the helpful comments and response below, I believe the following is a valid explanation:
(1) $\{\omega \in \Omega: X(\omega) \in A\} = \{\omega \in \Omega: X(\omega) \in A, Y(\omega) \in \mathbb{R}\}$, so the probability of these sets are equal.
(2) Definition of joint probability distribution measure
(3) Definition of joint probability density
(4) Tonelli's theorem
(5) One goal is to show $P(X \in A) = \int_A g(x) d\lambda(x)$ for some $g: \mathbb{R} \to \mathbb{R}$.  Certainaly $g(x) = \int_{\mathbb{R}} f_{X,Y} d\lambda(y) : \mathbb{R} \to \mathbb{R}$, and since densities are a.s. unique, $g$ is indeed the density of $X$; i.e. $f_X = g$.
Finally, by the definition of a continuous random variable, $X$ is continuous.
 A: Proof: Consider $X$ and let $A \in \mathcal{B}$ where $\mathcal{B}$ is the Borel sigma algebra on $\mathbb{R}$.  Then
\begin{align}
P(X \in A) & = P(X \in A, Y \in \mathbb{R}) &\qquad (1)\\
& = P(\mathbf{X} \in A \times \mathbb{R}) &\qquad (2) \\
& = \int_{A \times \mathbb{R}} f(x,y) d\lambda^2 &\qquad (3) \\
& = \int_A \int_{\mathbb{R}} f(x,y) d\lambda(y) d\lambda(x) &\qquad (4) \\
& = \int_A f(x) d\lambda(x). &\qquad (5)
\end{align}
explaining each line:
(1)$\{\omega \in \Omega |X(\omega) \in A\} =\{\omega \in \Omega |X(\omega) \in A, Y(\omega \in \mathbb{R})\} $ $ \Bbb{P}(X \in A) = \Bbb{P}(\{\omega \in \Omega |X(\omega) \in A\}) =\Bbb{P}(\{\omega \in \Omega |X(\omega) \in A, Y(\omega \in \mathbb{R})\}) = \Bbb{P}(X\in A, Y \in \Bbb{R})  .$
(2) Definition of joint probability distribution measure
(3) Definition of joint probability density
(4) Tonelli's theorem
(5)Use the definition of $f(x) = \int_{\Bbb{R}} f(x,y) \, d\lambda(y)$
Now conclude that if $P(X \in A) = \int_A f(x) \, d\lambda(x)$ then $f(x)$ is the density of $X$
that $f(x)$ is a probability density follows from the definition of a probability density, any measurable function with the following property
$P(X \in A) = \int_A f(x) \, d\lambda(x)$
check it out: https://en.wikipedia.org/wiki/Probability_density_function#Formal_definition
