Supremum sum (Functions) [closed]

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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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
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.