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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 $2 < 3$ –  AnonymousCoward Sep 21 '11 at 6:46

2 Answers 2

up vote 1 down vote accepted

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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Does this follow from the fact that if we just assume that f(x) and g(x) are measurable, then: $f(x) > c - g(x)$, thus we can find a $q \in \mathbb{Q}$ such that $f(x) > q > c - g(x)$. This is just $\{f(x) > q\} \cap \{q > c-g(x)\}$. Take the union through all $q \in \mathbb{Q}$. –  emka Sep 21 '11 at 20:21
    
Furthermore it's the countable union of Borel sets. I believe each of those is already a Borel set. –  emka Sep 21 '11 at 20:38

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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Probably not helpful for EMKA. –  GEdgar Sep 21 '11 at 13:39
    
@GEdgar: What is the use of this comment of yours? –  Did Sep 21 '11 at 13:52

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