Theorem - If f and g are integrable on $[a,b]$, then $f+g$ is also integrable on $[a,b]$ with
$\int_a^bf+g=\int_a^bf+\int_a^bg$.
I am trying to write the proof as -
Consider $P={\{x_o,x_1,....,x_n}\}$ be a partition of $[a,b]$.
Define $m_i(f)=inf({\{f(x)|x\in [x_{i-1},x_i]}\}$,
$m_i(g)=inf({\{g(x)|x\in [x_{i-1},x_i]}\}$, and
$m_i(f+g)=inf({\{(f+g)(x)|x\in [x_{i-1},x_i]}\}$
$\forall i, m_i(f)+m_i(g)\leq f(x)+g(x),\;where\;x\in[x_{i-1},x_i]$
$\implies \forall i, m_i(f)+m_i(g)\leq (f+g)(x),\;where\;x\in[x_{i-1},x_i]$
$\implies \forall i, m_i(f)+m_i(g)\leq m_i(f+g)$
$\implies\sum_im_i(f)\Delta_i+\sum_im_i(g)\Delta_i\leq\sum_im_i(f+g)\Delta_i$
So, $\underline{S}(f,P)+\underline{S}(g,P)\leq \underline{S}(f+g,P),\;\forall\text{partitions}P\tag{1}$.
By $\epsilon$ criterion of infimum,
$\forall\epsilon>0,\;\exists f_i^*,g_i^*\;s.t.\;f_i^*<m_i(f)+\epsilon/2\;and \\ g_i^*<m_i(g)+\epsilon/2,\;\forall i$
$\implies\forall i,\; f_i^*+g_i^*<m_i(f)+m_i(g)+\epsilon$
$\implies m_i(f+g)<m_i(f)+m_i(g)+\epsilon$
By forcing principle,
$m_i(f+g)\leq m_i(f)+m_i(g)$
$\implies \underline{S}(f+g,P)\leq\underline{S}(f,P)+\underline{S}(g,P),\;\forall\text{partitions}P\tag{2}$
From $(1)$ and $(2)$,
$\underline{S}(f+g,P)=\underline{S}(f,P)+\underline{S}(g,P),\;\forall\text{partitions}P$
$\implies \underline{\int_a^b}f+g=\underline{\int_a^b}f+\underline{\int_a^b}g\tag{3}$
Similarly,$\overline{\int_a^b}f+g=\overline{\int_a^b}f+\overline{\int_a^b}g\tag{4}$
As $f$ and $g$ is integrable on $[a,b]$,
$\implies \underline{\int_a^b}f=\overline{\int_a^b}f\;and\;\underline{\int_a^b}g=\overline{\int_a^b}g$
Thus $(3)$ becomes,
$\underline{\int_a^b}f+g=\overline{\int_a^b}f+\overline{\int_a^b}g$
$\implies \underline{\int_a^b}f+g=\overline{\int_a^b}f+g$
and $\int_a^bf+g=\int_a^bf+\int_a^bg$
Q.E.D.
The proof looks correct to me.
But in book, $(3)$ is written as
$\underline{\int_a^b}f+g\geq\underline{\int_a^b}f+\underline{\int_a^b}g\tag{5}$
I am not able to understand how they come up with inequality? I get the strict equality in $(3)$.
Due to the above inequality, the steps of proof of the above theorem written in the book is different from that of mine.
My doubt is that how $(5)$ is true? I get $(3)$ in place of $(5)$?
Please clarify the doubt.
Your statement $f_i^{\ast}+g_i^{\ast}<m_i(f)+m_i(g)+\epsilon$ does not imply $m_i(f+g)<m_i(f)+m_i(g)+\varepsilon $ since the second statement is not true.
For instance take $f=-\chi_{[0,1]}$ and $g=-\chi_{[2,3]}$. Then $m_i(f+g)=-1$ but $m_i(f)+m_i(g)=-2$. This is due to the infimum being superadditive, because $f$ and $g$ might 'take' their infimal values at different points $x\in\mathbb{R}$.