let $a,b,$ and $c$ be real numbers with $a<c<b$ and function $f$ is integrable in intervals $[a,c]$ and $[c,b]$.
Show that the function f is integrable in $[a,b]$ and that
$\int_{a}^{b}f(x)dx = \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx$
What I have so far:
In the intervals $[a,c]$ and $[c,b]$, the function f is integrable. That means the Riemann sum in each interval converges to a real number $I_{i}$, where $i=1,2$, which represents the limit as the absolute value of partition converges to $0$.
How do I combine the 2 aforementioned Riemann sums so that there is a real number $I_{3}$, such that the Riemann sum in $[a,b]$ converges to $I_{3}$?
Since $f$ is integrable, by definition, \begin{align}I_1+I_2&=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx=\inf_{a=p_0<p_1<\cdots p_k=c}\sum_{i=1}^k \sup_{[p_{i-1},p_i]}f+\inf_{c=q_0<q_1<\cdots p_n=b}\sum_{i=1}^n \sup_{[q_{i-1},q_i]}f.\\ I_3&=\int_{a}^{b}f(x)dx=\inf_{a=t_0<t_1<\cdots t_m=b}\sum_{i=1}^k \sup_{[t_{i-1},t_i]}f. \end{align}
In particular, taking $m=k+n$, $t_i:=p_{i}$ for $i=0,\dots k$ and $t_{k+i}=q_{i}$ for $i=0,\dots n$, we have that any partition of $[a,c]\cup[c,b]$ is a partition of $[a,b]$, hence $I_3\le I_1+I_2$, because it is the infimum of a larger set.
To prove $I_3\ge I_1+I_2$, it suffices to think analogously, but using the fact that $$I_3=\int_{a}^{b}f(x)dx=\sup_{a=t_0<t_1<\cdots t_m=b}\sum_{i=1}^k \inf_{[t_{i-1},t_i]}f.$$