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I know that , for any bounded sets $A,B\subset \mathbb{R}$, $\sup A + \sup B = \sup (A+B)$.

But what if $A,B$ are not arbitrary, but functions?

Let $f_i:[a,b]\rightarrow \mathbb{R}$ be bounded functions and $\alpha:[a,b]\rightarrow \mathbb{R}$ be a monotonically increasing function and $P$ be any partition of $[a,b]$.

How do i prove that $L(P,f_1,\alpha) + L(P,f_2,\alpha) ≦ L(P,f_1 + f_2, \alpha)$?

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closed as not a real question by Did, martini, Thomas, froggie, rschwieb Nov 25 '12 at 20:00

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

    
I don't know how to close this post.. Sombody do it for me please. i have no idea why i misunderstood Lower bound to be a supremum and took about an hour to prove this :( –  Katlus Nov 25 '12 at 16:27
    
You should have a "delete" option right below the (real-analysis) tag. –  kahen Nov 25 '12 at 16:51
    
@kahen I don't.. I'm a unregistered user and now i want to change it to be registered, but don't know how.. –  Katlus Nov 25 '12 at 16:55
    
This might be a duplicate question, but without knowing what $L$ is, it's hard to tell. Is $L(P,f,\alpha)$ supposed to be a lower Riemann sum approximating $\int_a^b f\,\mathrm{d}\alpha$? –  robjohn Nov 25 '12 at 17:31
    
@Katlus: I don't know exactly what an unregistered user sees, but click on "my logins" on your profile page, and see if you can add a login or register in some other way. –  robjohn Nov 25 '12 at 17:53
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1 Answer 1

I think this question is asking how to show something like $$ \sup_{x\in A}(f(x)+g(x))\le\sup_{x\in A}f(x)+\sup_{x\in A}g(x)\tag{1} $$ This is an instance of the fact that the supremum over a set is no smaller than the supremum over a subset. The left hand side of $(1)$ is $$ \sup_{\substack{x,y\in A\\x=y}}(f(x)+f(y))\tag{2} $$ whereas the right hand side of $(1)$ is $$ \sup_{x,y\in A}(f(x)+f(y))\tag{3} $$ The set of $x$ and $y$ being considered in $(2)$ is a subset of the $x$ and $y$ being considered in $(3)$, so $(1)$ follows.

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