Show that f+g is integrable and find its value

[evinda]

Hey again! (Blush)

Let $f,g:[a,b] \to \mathbb{R}$ integrable.Then,show that $f+g$ is integrable and $\displaystyle \int_a^b{(f+g)}=\int_a^b{ f } +\int_a^b{ g }$.

That's what I have tried:

We consider the partition $P=\{a=t_0<t_1<...<t_n=n\}$ of $[a,b]$.

$\forall t \in [t_k,t_{k+1}]: \sup f([t_k,t_{k+1}]) \geq f(t) \text{ and } \sup g([t_k,t_{k+1}]) \geq g(t)$

So, $\sup f([t_k,t_{k+1}]) + \sup g([t_k,t_{k+1}]) \geq f(t)+g(t) \Rightarrow \sup f([t_k,t_{k+1}]) + \sup g([t_k,t_{k+1}]) \geq \sup(f+g)[t_k,t_{k+1}]$

So,we conclude that: $U(f+g,P) \leq U(f,P)+U(g,P) \text{ and } L(f+g,P) \geq L(f,P)+L(g,P)$

Let $\epsilon'>0$
Since, $f,g$ are integrable,there are partitions $P_1,P_2$ of $[a,b]$ such that:

$U(f,P_1)-L(f,P_1)< \epsilon'$

$U(g,P_2)-L(g,P_2)< \epsilon'$

We take $P=P_1 \cup P_2$

So,we get: $U(f,P)-L(f,P)<\epsilon' \text{ and } U(g,P)-L(g,P)<\epsilon'$

$U(f+g,P)-L(f+g,P) \leq (U(f,P)-L(f,P))+(U(g,P)-L(f,P)) < 2\epsilon'=\epsilon$ (we pick $\epsilon'=\frac{\epsilon}{2}$)

So,$f+g$ is integrable.$\displaystyle \int_a^b{(f+g)} \leq U(f+g,P) \leq U(f,P)+U(g,P) < L(f,P)+L(g,P)+ \epsilon < \underline{\int_a^b f}+ \underline{\int_a^b g}+ \epsilon=\int_a^b f + \int_a^b g+\epsilon, \forall \epsilon \Rightarrow \int_a^b {f+g} \leq \int_a^b f + \int_a^b g$

$\displaystyle \int_a^b{(f+g)} \geq L(f+g,P) \geq L(f,P)+L(g,P)> U(f,P)+U(g,P)-\epsilon > \overline{\int_a^b f}+\overline{\int_a^b g}-2\epsilon=\int_a^b f + \int_a^b g-\epsilon, \forall \epsilon \Rightarrow \int_a^b {f+g} \geq \int_a^b f + \int_a^b g$

Therefore, $\displaystyle \int_a^b {(f+g)}=\int_a^b f +\int_a^b g.$

Is it right or am I wrong??

 

[stainburg]

It's OK!

 

[evinda]

It's OK!

Ok,thanks! :)