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Durrett has a theorem that says: if $X_1, X_2, ..., X_n$ are random variables then $X_1 + X_2 + ... + X_n$ are also random variables.

My issue is how to show that $F((x_1, x_2, ... , x_n)) = \sum_{i = 1}^{n} x_i$ is measurable?

Durrett says ${x_1 + x_2 + ... + x_n < a}$ is an open set $(-\infty, a)$ so then it's measurable and this is what I don't understand.

I appreciate your help.

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up vote 1 down vote accepted

Let $F: \mathbb{R}^n \to \mathbb{R}$. Then $F$ is measurable if $F^{-1}(-\infty,a) \in \mathcal{B}(\mathbb{R}^n)$ for all $a \in \mathbb{R}$ (where $\mathcal{B}(\mathbb{R}^n)$ denote the borel sets on $\mathbb{R}^n$).

Since $(x_1,\ldots,x_n) \to F(x_1,\ldots,x_n):=\sum_{i=1}^n x_i$ is a continuous mapping, we know that pre-images of open sets are open. Hence $F^{-1}(-\infty,a)$ is a open set. Since $\mathcal{B}(\mathbb{R}^n)$ is generated by the open sets (and therefore in particular contains all open sets) we conclude $F^{-1}(-\infty,a) \in \mathcal{B}(\mathbb{R}^n)$

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First show that if $f_1$ and $f_2$ are measurable, then so is $f_1 + f_2$. This is so since $$\{x: f_1 + f_2 < a\} = \bigcup_{r \in \mathbb{Q}} \left( \underbrace{\{x: f(x) < r\}}_{\text{Measurable}} \cap \underbrace{\{x: g(x) < a-r\}}_{\text{Measurable}} \right)$$ Hence, $f_1 + f_2$ is measurable since the above set is a countable union/intersection of measurable sets. Now use induction to conclude what you want.

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