TITLE: Why does continuity imply $P(X_i = X_j) = 0$, without integrals
QUESTION [5 upvotes]: In the book A probability path, at chapter 4 integrals were not yet defined. At 4.3.1, it says

Let $\{X_n, n\geq 1\}$ be iid (real random variables) with common distribution function
$F(x)$. The continuity of $F$ implies $$P[X_i = X_j] = 0$$

I'm guessing they mean for some $i\neq j$. Intuitively sounds like that set is going to be a curve and therefore of null measure in $\mathbb{R}^2$. Is there a simple way to prove this without using integrals?

REPLY [5 votes]: Let $X$ and $Y$ be i.i.d. with continuous CDF $F(x) = P[X \leq x]$ for all $x \in \mathbb{R}$. We want to show $P[X=Y]=0$.
Fix $m$ as a postive integer. Partition $\mathbb{R}$ into $m$ disjoint subintervals $\{I_k\}_{k=1}^m$ such that
\begin{align*}
&\cup_{k=1}^m I_k = \mathbb{R}\\
&P[X \in I_k] = 1/m \quad \forall k \in \{1, ..., m\}
\end{align*}
This is possible because the CDF of $X$ is continuous.  Now
$$ \{X=Y\} \subseteq \cup_{k=1}^m \{X \in I_k\}\cap \{Y \in I_k\}$$
So
\begin{align}
P[X=Y] &\leq \sum_{k=1}^m P[\{X \in I_k\}\cap \{Y \in I_k\}]\\
&= \sum_{k=1}^m P[X \in I_k]P[Y \in I_k]\\
&=\sum_{k=1}^m (1/m)^2\\
&= 1/m
\end{align}
This holds for all positive integers $m$ and so
$$P[X=Y]=0$$