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Let $R \subset \Bbb{R}^n$ and $f,g$ be two integrable functions from $R\to \Bbb{R}$ , in a proof of a property of integrals Rudin writes the following: $$L(P,f)+L(P,g) \le L(P,f+g) \le U(P,f+g) \le U(P,f)+U(P,g)$$ where $P$ is any partition. The $\le$ between $L(P,f+g)$ and $U(P,f+g)$ is clear but why is it that $L(P,f)+L(P,g) \le L(P,f+g)$ and $U(P,f+g) \le U(P,f)+U(P,g)$ isn't it easy to see by expanding the sums to see that they are equal since $\sup(f+g)=\sup(f)+\sup(g)$ (and similarly for the $\inf$) ? Am I missing something or did Rudin just write it this way since it won't make a difference?

Note: Rudin's proof is over an interval, but the multivariable case is similar.

user10444
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1 Answers1

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You have made a small error. It is true that $\sup(f+g) \leq \sup f + \sup g$. The inequality may be strict though. (Can you come up with an example for yourself? Consider $f$ and $g$ which obtain their maxima at different places.)

nullUser
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  • I seem to have ignored the fact the $\sup$ is taken over each subinterval, what still confuses me though is: what is missing in the conditions for the relation $\sup(f+g)=\sup(f)+\sup(g)$ to hold, since it only requires the functions to be bounded which is true here. Here is a proof : http://math.stackexchange.com/questions/4551/how-can-i-prove-supab-sup-a-sup-b-if-ab-ab-mid-a-in-a-b-in-b – user10444 Mar 21 '13 at 19:52
  • You are confusing yourself with notation, $\sup(f+g)$ and $\sup(A+B)$ mean different things. We have $\sup_X(f+g) = \sup{f(x)+g(x) : x \in X }$ while $\sup(A+B) =\sup{a+b: a \in A, b \in B}$. It is true that $\sup(f(X) + g(X)) = \sup f(X) + \sup g(X)$ where $f(X),g(X)$ denote the images (ranges) of $f,g$ respectively. – nullUser Mar 21 '13 at 20:09
  • In Particular, $(f+g)(X) \neq f(X)+g(X)$. – nullUser Mar 21 '13 at 20:11
  • Oh I get it now, thank you! – user10444 Mar 21 '13 at 20:32