Suppose $f$ is bounded on $[a,c]$ and $aSuppose $f$ to be a bounded function on the interval $[a,c]$ and where $a<b<c$. Prove that
$$
\overline{\int_{a}^{c}}f = \overline{\int_{a}^{b}}f + \overline{\int_{b}^{c}}f
$$
I've been trying to directly prove it by assigning two partitions on each interval and making one which refines both - got further than what I was asked to prove.
Also tried on the contrary but the only  idea I had was to say suppose:
$$
\overline{\int_{a}^{c}}f \not= \overline{\int_{a}^{b}}f + \overline{\int_{b}^{c}}f
$$
then if $f$ was integrable the above was absurd but couldn't expain why should assuming something that isn't given could work.
much appreciate for any help
 A: Take arbitrary partitions $P_{ab} : a = x_0 < x_1 < \ldots < x_n = b$ and  $P_{bc} : b = y_0 < y_1 < \ldots < y_m = c$ of the intervals $[a,b]$ and $[b,c]$, respectively.
Together they form a partition of  the interval $[a,c]$: $$P_{ac}: a = x_0 < \ldots< x_{n-1} < b < y_1 < \ldots < y_m = c$$
The upper Darboux sums of $f$ over these intervals satisfy
$$U(f,P_{ab}) + U(f, P_{bc}) =U(f, P_{ac}) \geqslant \inf \,\{U(f,P)\,|\, P \text{ is a partition of } [a,c]\} := \overline{\int_{a}^{c}}f $$
Since the partitions $P_{ab}$ and $P_{bc}$ are independent, we can take infima succcesively to argue
$$ U(f, P_{bc})\geqslant - U(f,P_{ab}) + \overline{\int_{a}^{c}}f  \implies \overline{\int_{b}^{c}}f \geqslant - U(f,P_{ab}) + \overline{\int_{a}^{c}}f\\ \implies U(f,P_{ab})  \geqslant -\overline{\int_{b}^{c}}f + \overline{\int_{a}^{c}}f  \implies  \overline{\int_{a}^{b}}f \geqslant -\overline{\int_{b}^{c}}f + \overline{\int_{a}^{c}}f  \implies\\   \boxed{\overline{\int_{a}^{b}}f +\overline{\int_{b}^{c}}f  \geqslant \overline{\int_{a}^{c}}f}   $$
To prove the inequality in the other direction take any partition $P_{ac}$ of $[a,c]$ and insert the point $b$ to obtain a refined partition $P'_{ac}$ which induces partitions $P_{ab}$ and $P_{bc}$ of $[a,b]$ and $[b,c]$, respectively. Since upper sums decrease under refinement we have
$$U(f, P_{ac})\geqslant U(f,P'_{ac}) = U(f,P_{ab}) + U(f, P_{bc}) \geqslant \overline{\int_{a}^{b}}f +\overline{\int_{b}^{c}}f$$
Taking the infimum over all partitions of $[a,c]$ we get
$$ \boxed{\overline{\int_{a}^{c}}f\geqslant \overline{\int_{a}^{b}}f +\overline{\int_{b}^{c}}f}   $$
Altogether, this proves
$$\overline{\int_{a}^{c}}f = \overline{\int_{a}^{b}}f +\overline{\int_{b}^{c}}f$$
