Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that

(i)	$X_1, Y_1, \, \ldots, \, X_N, Y_N : \Omega \to \mathbb{R}^D$ are each a D4383: Random euclidean real number on $P$

Let $\omega \in \Omega$ be an outcome such that $\sum_{n = 1}^N X_n(\omega) \neq \sum_{n = 1}^N Y_n(\omega)$. Then $X_n(\omega) \neq Y_n(\omega)$ for at least one $n \in 1, \ldots, N$, for otherwise the sums $\sum_{n = 1}^N X_n(\omega)$ and $\sum_{n = 1}^N Y_n(\omega)$ would be equal. Thus, by definition of a union, $\omega \in \bigcup_{n = 1}^N \{ X_n \neq Y_n \}$. Since $\omega \in \Omega$ was arbitrary, we have the inclusion \begin{equation} \left\{ \sum_{n = 1}^N X_n \neq \sum_{n = 1}^N Y_n \right\} \subseteq \bigcup_{n = 1}^N \{ X_n \neq Y_n \} \end{equation} Result R2090: Isotonicity of probability measure now implies \begin{equation} \mathbb{P} \left( \sum_{n = 1}^N X_n \neq \sum_{n = 1}^N Y_n \right) \leq \mathbb{P} \left( \bigcup_{n = 1}^N \{ X_n \neq Y_n \} \right) \end{equation} which is what was required to be shown. $\square$