How can I prove that the sum of two parts of an integral is always equal to the sum regardless of order?

I want to show that if $f$ is an integrable function on $[a,c]$, then the following statement is true:

$\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx=\int_{a}^{c}f(x)dx$

The twist is that this is true for any order of $a,b,$ and $c$ no matter which position they may have. But since we are assuming that the function itself is integrable on all parts, I think we can also use this:

$\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$

To start off my proof, I will begin with let $P$ be a partition of $[a,b]$ and $Q$ a partition of $[b,c]$. Then, $P \cup Q$ must be a partition of $[a,c]$. Then, $L(P,f) + L(Q,f) = L(P \cup Q,f)$ and also $U(P,f)+U(Q,f) = U(P \cup Q,f)$. In addition, it follows that for any partition $R$, $L(R,f) \leq L(R \cup \{b\},f) \leq U(R \cup \{b\},f) \leq U(R,f)$.

Thus, $\int_{a}^{c}f(x)dx = \inf_{R}U(R,f) =\inf_{R}U(R \cup \{b\},f)=\inf_{P \cup Q}U(P \cup Q,f) = \inf_{P}U(P,f)+\inf_QU(Q,f)=\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx$

I think the proof as is good but I am not sure how I can utilize the part where order does not matter to finish my proof. I do believe that we can use WLOG and only do it for one combination and say it is similar to all other possible combinations.

Solution 1:

Take the theorem to hold for $$a\leq b\leq c$$ in that order. Now consider $$a\leq c\leq b$$. In shorthand $$\int_{a}^{b} = \int_{a}^{c} + \int_{c}^{b}$$. $$\int_{a}^{b} = \int_{a}^{c} - \int_{b}^{c}$$ $$\int_{a}^{b} + \int_{b}^{c} = \int_{a}^{c}$$ as proposed