[Math] Sum of Riemann integrable functions

real-analysis

I am trying to study real analysis on my own and I need help proving that the sum of two riemann integrable functions is also riemann integrable.
The exact problem statement from the book:

Let $f$ and $g$ be bounded, Riemann integrable functions on $[a,b]$. Show that $f+g$ is Riemann integrable on $[a,b]$ with $\int\limits_a^b(f+g)=\int\limits_a^bf+\int\limits_a^bg$.

My attempt at the solution so far:

I am trying to prove this using the integrability criterion $U(f+g,P)-L(f+g,P)<\epsilon$, where $U$ and $L$ are the upper and lower sums of $f+g$ with respect to $P$.

Let $P_f$ and $P_g$ be partitions of $[a,b]$ that satisfy the condition. Fix $\epsilon >0$. By the integrability of $f$ and $g$:

$U(f,P_f)-L(f,P_f)<\frac{\epsilon}{2}$

$U(f,P_f)-L(f,P_f)<\frac{\epsilon}{2}$

$U(g,P_g)-L(g,P_g)<\frac{\epsilon}{2}$

Let $P = P_f\bigcup P_g$. Since $P$ is a refinement of both $P_f$ and $P_g$ we have

$L(f,P_f)\leq L(f,P)$, $U(f,P)\leq U(f,P_f)$, $L(g,P_g)\leq L(g,P)$, $U(g,P)\leq U(g,P)$

Therefore,

$U(f,P)-L(f,P)<\frac{\epsilon}{2}$

$U(g,P)-L(g,P)<\frac{\epsilon}{2}$

Now I am stuck because I can't establish a relation between the upper sum of each function individually and the upper sum of both functions with respect to $P$ (Same for lower sums). Any advice?

Best Answer

Let $P$ be any partition of $[a,b]$. If $J$ is an interval in the partition then it is easy to see that $\sup_J(f+g)\leq \sup_Jf+\sup_Jg$ and $\inf_J(f+g)\geq \inf_Jf+\inf_Jg$. So if the partition $P$ is $a=x_0<x_1<...<x_n=b$ then:

$U(f+g,P)=\sum_{i=1}^n \sup_{[x_{i-1},x_i]}(f+g)\times (\Delta x_i)\leq \sum_{i=1}^n \sup_{[x_{i-1},x_i]}(f)\times (\Delta x_i)+\sum_{i=1}^n \sup_{[x_{i-1},x_i]}(g)\times (\Delta x_i)=U(f,P)+U(g,P)$

$L(f+g,P)=\sum_{i=1}^n \inf_{[x_{i-1},x_i]}(f+g)\times (\Delta x_i)\geq \sum_{i=1}^n \inf_{[x_{i-1},x_i]}(f)\times (\Delta x_i)+\sum_{i=1}^n \inf_{[x_{i-1},x_i]}(g)\times (\Delta x_i)=L(f,P)+L(g,P)$

Can you finish from here?