How to prove: if a function is integrable on $[a, b]$ then $\int_{a}^{c} f(x) + \int_{c}^{b}f(x)=\int_{a}^{b}f(x)$ So I'm stuck on this question, I do not really understand how to do it. I did the previous question which was apart of it (proving $f$ is integrable on $[a, c]$ and $[c, b]$, so I am wondering if someone can help me solve it, thanks! :)

Suppose the function $f : [a, b] \rightarrow \mathbb{R}$ is integrable on [a, b]. Prove that $f$ is integrable on $[a, c]$ and $[c, b]$ for every $c\in (a,b)$. Furthermore, $\int_{a}^{b}f(x) dx = \int_{a}^{c}f(x) dx + \int_{c}^{b}f(x) dx $.

Working
Since $f$ is integrable on $[a, b] \implies U(f, P)-L(f, P)<\epsilon$
Fix $\epsilon > 0$ 
Then $\exists P_1$ such that $\int_{a}^{c}f(x) dx=\int_{\underline{a}}^{c}f(x) dx<L(f, P_1)+\frac{\epsilon}{2}$
Also $\exists P_2$ such that $\int_{a}^{c}f(x) dx=\int_{a}^{\overline{c}}f(x) dx>U(f, P_2) - \frac{\epsilon}{2}$
Let $P=P_1\cup P_2$, then 
$U(f, P) - L(f, P)\leq U(f, P_2)-L(f, P_1)<\int_{a}^{c}f(x) dx + \frac{\epsilon}{2} - \int_{a}^{c}f(x) dx + \frac{\epsilon}{2}=\epsilon$
[c, b] is analagous.
Note
I just need help help with the second part of the question, however if you wanted to check my current working that would be appreciated as well, again, thanks! :)
 A: Using the integral Criterion:
Suppose for the sake of contradiction that $f$ is not integrable on either $[a,c]$ or $[c,b]$. Without loss of generality, say it is $[a,c]$. Then $\exists \epsilon_0$ for all partitions $P$ of $[a,c]$ such that $U(f,P)-L(f,P)\geq \epsilon_0$.
However, we know that $f$ is integrable on $[a,b]$, so there must be a partition $P_0$ of $[a,b]$ such that $U(f,P_0)-L(f,p_0)<\epsilon_0$.
However, consider the partition $P_1=P_0 \cap [a,c]$. $P_1$ is a subset of $P_0$ so by the refinement theorem
$U(f,P_1)-L(f,P_1)<U(f,P_0)-L(f,P_0)<\epsilon_0$.
However, this is a contradiction, because $P_1$ is a partition of $[a,c]$ and we are supposed to have that for any partition of $[a,c]$, $U(f,P)-L(f,P)\geq \epsilon_0$.
Using Lebesgue Theorem:
Lebesgue theorem states that a function is integrable if and only if the set $S=\{x:f \text{ is discontinuous at }x\}$ is a measure $0$.
Since $f$ is integrable on $[a,b]$ then the set $s_{f_{[a,b]}}$ which is the set of discontiniuities of $f$ on $[a,b]$ is of measure $0$. Since $[a,c]\subset [a,b]$ then the set of discontinuities of $f$ in $[a,c]$, namely $S_{f_{a,c}}\subset S_{f_{[a,b]}}$. Hence it is a measure $0$ (Why?) and thus $f$ is Riemann integrable on $[a,c]$. Same argument for $[c,b]$.
