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Let $\mathcal{E}=\{E_n : n\geq 1\}$ be a sequence of independent exponential random variables with mean 1. How can I see the set $E=\{E_n(\omega) : n\geq 1\}$ dense in $[0,\infty)$? I imagine one can make an argument about the cdf being discontinuous if there's an open set that doesn't contain any $E_n(\omega)$, but I don't know if this is even the right direction.

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Let $a <b$. Then $$P(E_n \notin (a,b) \, \forall n \in \{1,2,..,N\})=\prod_{k=1}^{N} P(E_n \notin (a,b))$$ $$ =\prod_{k=1}^{N} (e^{-a}-e^{-b})=(e^{-a}-e^{-b})^{N} \to 0$$ as $N \to \infty$. Hence, $P(E_n \notin (a,b) \, \forall n )=0$. Taking union over all rational values of $a$ and $b$ we see that there is one null set $E$ such that $\omega \notin E$ implies $(E_n(\omega))$ intersects every open interval in $\mathbb R$.

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