24
$\begingroup$

Suppose $f$ and $g$ are Lebesgue measurable, we want to show $f+g$ is measurable. So, the hint is to consider the continuous functions $F : \mathbb{R}^2 \to \mathbb{R} $ given by $h(x) = F(f ,g ) $. If we can show $F$ Is measurable, then Taking $F = f +g $ would solve our problem.

In other words, I want to show that the set $R = \{ (f,g) : F(f,g) > a $ } is lebesgue measurable.. But this set is just a rectangle in the plane. And since $F$ is continuous, then $R$ must be open, and hence a union of open rectangles which are measurable and hence $R$ must be measurable. Is this a correct approach to the problem? Can someone help me to make this formal? thanks

$\endgroup$
46
$\begingroup$

This is a method I used in my analysis class. Note that $f(x) + g(x) < t$ iff $f(x) < t-g(x)$ iff there exists a rational number $r$ such that $f(x) < r < t-g(x)$.

Therefore $\{x : f(x) + g(x) < t\} = \bigcup_{r\in\Bbb Q} [f^{-1}((-\infty, r)) \cap g^{-1}((-\infty, t-r))]$.

Both the sets being intersected in the union are measurable sets. Hence the set on the left is also measurable, meaning that $f+g$ is measurable.

$\endgroup$
  • $\begingroup$ Could you please expalain, why we considering rational number 'r'. Can we take real number? and I think t is real? $\endgroup$ – prasad Feb 13 '14 at 9:27
  • 14
    $\begingroup$ We want to use a rational number so that our union is countable. A COUNTABLE union of measurable sets is measurable. I do not believe we can say as much for an uncountable union. $\endgroup$ – Vladhagen Feb 13 '14 at 23:51
  • 2
    $\begingroup$ For an uncountable union we can't even say a thing. Any unmeasurable set of reals is an uncountable union of points, each of which has measure zero. $\endgroup$ – user21820 Jul 21 '15 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.