Confusion with real numbers and random variables; Integration and Independence in Williams In David Williams' Probability with Martingales, $\exists$ this exercise.
Let $X_n$ be iid RVs with the same continuous dist function. Let $E_1 =  \Omega$ and for $n \geq 2, E_n = (X_n > X_m \forall m < n)$
Exercise: Prove that $E_i$'s are independent with $P(E_n) = 1/n$.
After loopy integration by parts, I was able to show that $P(E_n) = 1/n$.
For independence, I tried starting with something specific that will be generalized later on:
$P(E_2 \cap E_4) = P(E_2)P(E_4)$
No problems with RHS obviously. As for the LHS:
$E_2$ means $X_2 > X_1$
$E_4$ means $X_4 > X_3,X_2,X_1$
Now $P(E_2 \cap E_4) = P(X_2 > X_1, X_4 > X_3,X_2,X_1) $.
Here is where I am confused. My prof says that $P(E_2 \cap E_4) = P(X_2 > X_1, X_4 > X_3,X_2) $ since the $X_4 > X_1$ part is implied from $X_2 > X_1$ and $X_4 > X_2$ .
Is that correct? I tried doing the integration and I was able to get 1/8 without removing $X_4 > X_1$. Removing it gave me 1/6, I think.
My main confusion: I have a feeling that the $X_4 > X_1$ is not removed because $X_4$ is now treated as a real number i.e. $X_m$ in the original event is a real number. After all, it just ends up being in the upper bound of the integral when integrating the joint probability/cumulative distribution functions. Is that wrong?
Edit: I figured out what I meant to say. I don't think $X_4 > X_1$ is eliminated since one $X_2$ is a RV and another $X_2$ is a real number.
 A: First, $P(E_n)=1/n$ simply by symmetry, because $E_n$ happens when $\max\{X_1,X_2,\ldots,X_n\}$ is realized with $X_n$. Since the common distribution is continuous there is almost surely no ex aequo and $\max\{X_1,X_2,\ldots,X_n\}$ has as much chances to be realized with each $X_k$ with $1\leqslant k\leqslant n$, hence the result.
Likewise, the sigma-algebra $\mathcal E_n=\sigma(E_k;1\leqslant k\leqslant n)$ contains only information about the ordering of the sample $(X_k)_{1\leqslant k\leqslant n}$ while the conditional probability of $E_{n+1}$ conditionally on  $\mathcal E_n$ depends only on $\max\{X_1,X_2,\ldots,X_n\}$, which is independent of this ordering. 
Thus,  the event $E_{n+1}$ and the sigma-algebra $\mathcal E_n$ are independent, that is, $E_{n+1}$ is independent of $(E_k)_{1\leqslant k\leqslant n}$, QED.
Edit: (To answer a question asked in a comment.) 
By definition, $E_2 \cap E_4 = [X_2 > X_1, X_4 > X_3,X_4\gt X_2,X_4\gt X_1] $ hence, indeed $E_2 \cap E_4 = [X_2 > X_1, X_4 > X_3,X_4\gt X_2] $ since one omits only the condition $X_4\gt X_1$, which is a consequence of the conditions $X_2\gt X_1$ and $X_4\gt X_2$ one keeps. These are identities between events, which imply identities between their respective probabilities. There are no "real numbers" here, only random variables.
Edit-edit: Two formulas for $P(E_2 \cap E_4)$ as multiple integrals are proposed in a comment. Only the first one seems to make sense since the second one integrates a function of $(x_1,x_2,x_3,x_4)$ with respect to some mysterious ($5$-dimensional?) measure $dx_1dx_1dx_2dx_3dx_4$. The first formula suggested is: $$P(E_2 \cap E_4) = \int_{-\infty}^{\infty} \int_{-\infty}^{x_4} \int_{-\infty}^{x_4} \int_{-\infty}^{x_2} f(x_1)f(x_2)f(x_3)f(x_4)\mathrm dx_1\mathrm dx_2\mathrm dx_3\mathrm dx_4.$$
Performing the simple integrations involved in the RHS from the innermost to the outermost, one gets successively the results $$f(x_2)f(x_3)f(x_4)F(x_2),\qquad\frac12f(x_3)f(x_4)F(x_4)^2,\qquad\frac12f(x_4)F(x_4)^3,$$ hence the result is $$\int_{-\infty}^{\infty}\frac12f(x_4)F(x_4)^3\mathrm dx_4=\left.\frac18F(x_4)^4\right|_{-\infty}^{\infty}=\frac18.$$
