Why do IID random variables with a continuous distribution function have distinct values from one another a.s? Let $\{X_i\}$ be a sequence of IID random variables with a continuous distribution function $F_X$. Then, why are the values of $X_i$'s distinct a.s?
I can see that for 'each' $X_i$, the measure of $X^{-1}(a)$ is zero for all $a \in \mathbb{R}$, but I cannot find any relation among $X_i$'s.
Could anyone please help me?
 A: For a continuous random variable $X$ we have $P(X=x)=0$ for all $x$. Now, by independence, $P(X_i=X_j)=\int P(X_i=x) dF_{X_j} (x)=\int 0 dF_{X_j} (x)=0$ if $i \neq j$.
We can also say that $P(\bigcup_{i \neq j} \{X_i =X_j\})=0$
A: It suffices to show the result for two iid r.v.'s $X_1, X_2$ with continuous cdf $F_X$.
For each $q \in \Bbb{Q} \cap (0, 1)$, let $r_q := F_X^{-1}(q)$, $r_0 := -\infty$, $r_1 := \infty$. We can cover the diagonal $\{ (x, x): x \in \Bbb{R} \} \subset \Bbb{R}^2$ by the rectangles of the form $\left( r_{(i-1)/n}, r_{i/n} \right)^2$ (minus finitely many points, having zero probability).
Since the probability that $X_1 = X_2$ is the probability that the random point $(X_1, X_2)$ is in the diagonal, for all $n \geq 0$,
\begin{align*}
P(X_1 = X_2) & \leq \sum_{i=1}^n P \left( (X_1, X_2) \in \left( r_{(i-1)/n}, r_{i/n} \right)^2 \right) \\
 &= \sum_{i=1}^n \left[ F(r_{i/n}) - F(r_{(i-1)/n})\right]^2 \\
 &= \sum_{i=1}^n \left[ \frac{i}{n} - \frac{i-1}{n} \right]^2 \\
 &= \sum_{i=1}^n \frac{1}{n^2} \\
 &= \frac{1}{n},
\end{align*}
and since $n$ was arbitrary, $P(X_1 = X_2) = 0$.
