Can anyone explain the convolution of independent random variables? Specifically, suppose that $X_1$ and $X_2$ are independent and $f$ is a Borel-measurable function s.t. $f(x_1, x_2)=\begin{cases}
1, & \text{if } x_1+x_2 \leq x\\
0, & \text{otherwise}
\end{cases}$ and $\mu$ is a product measure for $X_1$ and $X_2$.
Why do the following steps hold?

\begin{align}
\int_\Omega f(X_1,X_2)\,d\mathscr P & = \iint\limits_{\mathscr R^2} f(x_1,x_2) \, \mu^2(dx_1,dx_2) \\[6pt]
& = \int_{\mathscr R^1} \mu_2(dx_2) \int_{\mathscr R^1} f(x_1,x_2)\,\mu_1(dx_1) \\[6pt]
& = \int_{\mathscr R^1} \mu_2(dx_2) \int_{(-\infty,x-x_2]} \mu_1(dx_1) \\[6pt]
& = \int_{-\infty}^\infty dF_2(x_2) F_1(x-x_2).
\end{align}
 A: The first step is probably the hardest.  It's the "change of variables" theorem of measure theory.  If $f \in L^{1}(\Omega)$, then
\begin{equation}
\int_{\Omega} f(X_{1}(\omega),X_{2}(\omega)) \, \mathbb{P}(d\omega) = \int_{\mathbb{R}^{2}} f(x_{1},x_{2}) \, \mu^{2}(x_{1},x_{2}).
\end{equation}  Here $\mu^{2}$ is the measure you obtain when you "pushforward" $\mathbb{P}$ by $(X_{1},X_{2})$, which is a fancy way of saying $$\mu^{2}(A) = \mathbb{P}\{\omega \, \mid \, (X_{1}(\omega),X_{2}(\omega)) \in A\}$$ whenever $A \subseteq \mathbb{R}^{2}$.  Alternatively, if you're familiar with distribution functions, then $\mu^{2}$ is the measure corresponding to the distribution function of $(X_{1},X_{2})$.  Either way, if you're not familiar with the change of variables theorem, say so and I'm sure people on here can help out.
The rest is simpler.  We know
\begin{align*}
\mu^{2}(A_{1} \times A_{2}) &= \mathbb{P}\{\omega \, \mid \, X_{1}(\omega) \in A_{1}, \, \, X_{2}(\omega) \in A_{2}\} \\ 
&= \mathbb{P}\{X_{1} \in A_{1}\} \mathbb{P}\{X_{2} \in A_{2}\} \\
&= \mu_{1}(A_{1}) \mu_{2}(A_{2})
\end{align*} by independence.  (I dropped some $\omega$'s to eliminate unnecessary symbology.)  It follows by e.g. the Hahn-Caratheodory theorem that $\mu^{2}$ is the product of $\mu_{1}$ and $\mu_{2}$.  In particular, in the integral sign, we can replace $\mu^{2}(dx_{1},dx_{2})$ with $\mu_{1}(dx_{1}) \mu_{2}(dx_{2})$.   
Finally, applying Fubini's theorem (which is no problem since $f$ here is non-negative), 
\begin{align*}
\int_{\Omega} f(X_{1}(\omega),X_{2}(\omega)) \, \mathbb{P}(d \omega) &= \int_{\mathbb{R}^{2}} f(x_{1},x_{2}) \, \mu_{1}(dx_{1}) \mu_{2}(dx_{2}) \\
&= \int_{-\infty}^{\infty} \left(\int_{-\infty}^{\infty} \mathbf{1}_{\{x_{1} + x_{2} \leq x\}} \mu_{1}(dx_{1})\right) \mu_{2}(dx_{2}) \\
&= \int_{-\infty}^{\infty} \left(\int_{-\infty}^{x - x_{2}} \mu_{1}(dx_{1})\right)\mu_{2}(dx_{2}) \\
&= \int_{-\infty}^{\infty} F_{1}(x - x_{2}) \mu_{2}(dx_{2}).
\end{align*} 
(I replaced $f$ with the fancy $\mathbf{1}$ to help us figure out what the bounds of integration should be.)  Here I used the fact that $F_{1}$ is the distribution function of $\mu_{1}$, which just means $$F_{1}(y) = \mu_{1}(-\infty,y] = \int_{-\infty}^{y} \mu_{1}(dx_{1}).$$  Finally, we can replace $\mu_{2}(dx_{2})$ with $dF_{2}(x_{2})$ (or I would prefer $F_{2}(dx_{2})$), but either way it's just notational.  This is just the inverse of what I wrote above; it goes back to the correspondence between distribution functions and probability measures (on $\mathbb{R}$).  
A: The random variables $X_1$ and $X_2$ are functions from $\Omega$ into $\mathbb R$, and for Borel subsets $A$ of the plane $\mathbb R^2$, the mapping
$$
A \mapsto \mathscr P\left\{ \omega\in \Omega : (X_1(\omega), X_2(\omega)) \in A \right\}
$$
is a probability measure on the set of all Borel subsets of $\mathbb R^2$.  For the moment let us call that measure $\nu$; then we have
$$
\nu(A) = \mathscr P\left\{ \omega\in \Omega : (X_1(\omega), X_2(\omega)) \in A \right\}. \tag 1
$$
That is the definition of the measure $\nu$.  From this we wish to deduce that
$$
\int_\Omega f(X_1(\omega),X_2(\omega)) \, d\mathscr P(\omega) = \iint\limits_{\mathbb R^2} f(x_1,x_2)\, d\nu(x_1,x_2). \tag 2
$$
Any proof that $(2)$ is a consequence of $(1)$ would be no different from a proof that
$$
\int_\Omega g(X_1(\omega)) \, d\mathscr P(\omega) = \int\limits_{\mathbb R} g(x_1) \, d\mu(x_1)
$$
where $\mu$ is defined in a way analogous to $(1)$.  Maybe I'll return to the deduction of $(2)$ from $(1)$ tomorrow; I'll probably be able to say it more efficiently then.
We are told next that $\nu=\mu^2$, the product measure.  That implies both that $X_1,X_2$ are independent and that they are identically distributed, since the distribution of either of them is $\mu$.
Next we have the equality
$$
\iint\limits_{\mathbb R^2} f(x_1,x_2) \, \mu^2(dx_1,dx_2) = \int_{\mathbb R^1} \mu_2(dx_2) \int_{\mathbb R^1} f(x_1,x_2)\,\mu_1(dx_1).
$$
This says that a double integral, i.e. an integral with respect to a product measure, is equal to a corresponding iterated integral.  Fubini's theorem say that that is true when the integral of the absolute value is finite.  Since the measure $\mu^2$ assigns measure $1$ to all of $\mathbb R^2$, the remaining question of whether that integral is finite is just a question about the function $f$, and we know $f$ is bounded, and that's all we need to draw that conclusion.  Or we could use Tonelli's theorem instead of Fubini's theorem. Tonelli would give us this equality if $f\ge 0$ everywhere even if its integral were infinite.
Next we have the equality
$$
\int_{\mathbb R^1} \mu_2(dx_2) \int_{\mathbb R^1} f(x_1,x_2)\,\mu_1(dx_1) = \int_{\mathbb R^1} \mu_2(dx_2) \int_{(-\infty,x-x_2]} \mu_1(dx_1)
$$
This holds because
$$
\int_{\mathbb R^1} f(x_1,x_2)\,\mu_1(dx_1) = \int_{(-\infty,x-x_2]} \mu_1(dx_1),
$$
and that comes immediately from the definiton of $f$.
The very last equality comes from the definition of cumulative probability distribution functions.
A: $$\begin{align}
\iint_\Omega f(X_1, X_2)\operatorname d \mathscr P =&~ \iint_{\Bbb R^2} f(x_1,x_1)\,\mu^2(\operatorname d x_1, \operatorname d x_2) && \text{by definition of probability measure}
\\ =&~ \int_\Bbb R\color{navy}{\int_\Bbb R f(x_1, x_2)\mu_1(\operatorname d x_1)}\mu_2(\operatorname d x_2) && \text{by independence}
\\ = & ~ \int_\Bbb R\color{navy}{\int_{(-\infty;x-x_2]} \mu_1(\operatorname d x_1)}\mu_2(\operatorname d x_2) && \text{because}~f(x_1,x_2) {:=} \mathbf 1_{x_1\leqslant x-x_2}
\\ = & ~ \int_\Bbb R \color{navy}{F_1(x-x_2)}\operatorname d F_2(x_2) && \text{by definition of CDF}
\end{align}$$
A: Recall conditional expectation:
$$E(X) = E(E(X\mid\theta))$$
Condition on the value of $x_2$ (the last equality follows from independence): 
$$P(x_1 + x_2 < a) = P(x_1 < a - t : x_2 = t) P(x_2 = t) = P(x_1 < a - t) P(x_2 = t)$$
Now take expectation over $x_2$ to get rid of $t$ and you get
$$P(x_1 + x_2 < a) = \int_R F_1(a-x_2)dF_2(x_2).$$
Intuitively, for any value of $x_2=t$ the probabilty is the product of $F(t)$ and $F(x-t)$ and we need to take the average (probability-weighted) over all possible values of $t$.
