Showing that a function f+g is Borel measurable

Asked May 12 '23 at 08:43

Active May 12 '23 at 09:02

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I'm working on the following problem. I have found some definitions, that I think I can use. But I'm unsure how to implement them logically, so hope someone can help me proceed.

Problem:

Let f : R → R be a function such that f(x) ≤ f(y) for all x ≤ y, and let g : R → R be a function such that g(x) ≥ g(y) for all x ≤ y. Show that the function f + g is Borel measurable.

My attempt:

Using the proposition (X, A) is a measurable space, f , g ∈ M. Then f ± g are also measurable

Using the concept of monotiniticy since we have:

f(x) ≤ f(y) for all x ≤ y and g(x) ≥ g(y) for all x ≤ y

edited May 12 '23 at 09:02

Jean-Claude Arbaut

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asked May 12 '23 at 08:43

User4546

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Indeed, it suffices to show that $f$ and $g$ are both measurable. This follows from monotonicity as you can see here. – jakobdt May 12 '23 at 08:49
Thank you. So If I simply show that the functions are monotone (f(x) ≤ f(y) for all x ≤ y and g(x) ≥ g(y) for all x ≤ y) then they are Borel measurable? – User4546 May 12 '23 at 09:05
So, it is basically this I have to apply?: "If f is increasing, the set {x:f(x)>a} is an interval for all a, hence measurable." – User4546 May 12 '23 at 09:08
Question with same title – Тyma Gaidash May 12 '23 at 18:04

Showing that a function f+g is Borel measurable

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