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We say a function f is Borel measurable if for all $\alpha \in \mathbb{R}$ the set $\{x \in \mathbb{R} : f(x) > \alpha \}$ is Borel.

If f, g are Borel measurable, then f+g is Borel measurable.

First since f and g are Borel measurable then $\{x \in \mathbb{R}: f(x) > \alpha \}$ and $\{x \in \mathbb{R}: g(x) > \alpha \}$.

So f+g being Borel measurable, then I need to prove that $\{x \in \mathbb{R}: f(x)+g(x) > \alpha \}$ is Borel.

Isn't the last set just a union of $\{x \in \mathbb{R}: f(x) > \alpha \}$ and $\{x \in \mathbb{R}: g(x) > \alpha \}$ which are individually Borel measurable (I've proven this).

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    2+2 > 3 but 2 < 32011-09-21

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HINT: Write $\{x \in \mathbb{R}: f(x)+g(x) > \alpha \}$ as a countable union of sets easily seen to be Borel sets.

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    Furthermore it's the countable union of Borel sets. I believe each of those is already a Borel set.2011-09-21
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You might try to show first that $(f,g)$ is a Borel function from $\mathbb R$ to $\mathbb R^2$ and then that, for every $\alpha$, $\{(u,v)\in\mathbb R^2\mid u+v>\alpha\}$ is a Borel subset of $\mathbb R^2$.

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    @GEdgar: What is the use of this comment of yours?2011-09-21