Prove $\int_{a}^{c} f + \int_{c}^{b} f = \int_{a}^{b} f$ 
In terms of Calculus, it seems intuitive. Not sure how to formalize a proof.
Pf
Let $\epsilon > 0$ be given.
Since $ \int_{a}^{b} f$ exists, $\bar{S}(f;P(\epsilon)) - \underline{S}(f;P(\epsilon))< \epsilon$
Since $\int_{a}^{c} f$ exists, $\bar{S}(f;P_1(\epsilon)) - \underline{S}(f;P_1(\epsilon))< \frac{\epsilon}{2}$
Since $\int_{c}^{b} f$ exists, $\bar{S}(f;P_2(\epsilon)) - \underline{S}(f;P_2(\epsilon))< \frac{\epsilon}{2}$
We have $P_1 \cup P_2 = P$
Then $\inf\{\bar{S}(f;P|P$ of $[a,b]\} \leq \inf\{\bar{S}(f;P_1|P_1$ of $[a,c]\} + \inf\{\bar{S}(f;P_2|P_2$ of $[a,b]\}$
Thus $\int_{a}^{c} f + \int_{c}^{b} f \geq \int_{a}^{b} f$
Similarly the supremum shows$\int_{a}^{c} f + \int_{c}^{b} f \leq \int_{a}^{b} f$
Thus $\int_{a}^{c} f + \int_{c}^{b} f = \int_{a}^{b} f$
Not sure how to show this holds regardless of the relative size of a,b, and c.

I am not sure what you mean by sizes of $a, b, c$. I assume you mean the values that they take. In this case note that you can assume $a \le c \le b$ (all other cases are treated similarly) and you find: $$\int_{a}^{c}fdx = \int_a^bf1_{[a,c)}dx$$And: $$ \int_c^b fdx= \int_a^bf1_{[c,b)}dx$$
So the conclusion follows by additivity of the integral.
