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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).

emka
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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.

GEdgar
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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
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    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
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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 $\mathbb R^2$.

Did
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