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


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.

  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ This works only when f and g are finite a.e? $\endgroup$ – blue boy Feb 26 '20 at 0:49

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