Assume $f$ is Riemann integrable over $[a, b]$. Show that if $c \in [a, b]$ then $\int_a^c f(x)dx + \int_c^b f(x)dx = \int_a^b f(x)dx$
Note that that the version of the Riemann integral I'm using is defined through Darboux sums.
I rezlised that showing $$\int_a^c f(x)dx + \int_c^b f(x)dx = \int_a^b f(x)dx$$
was equivalent to showing that $$\inf \left\{\ U(f, P_{[a, c]}) \ | \ P_{[a, c]} \text{ is a partition of } [a, c]\right\} + \inf \left\{\ U(f, P_{[c, b]}) \ | \ P_{[c, b]} \text{ is a partition of } [c, b]\right\} = \inf \left\{\ U(f, P_{[a, b]}) \ | \ P_{[a, b]} \text{ is a partition of } [a, b]\right\}$$
because since $f \in R[a, b]$ and $[a, c], [c, b] \subseteq [a, b]$ it follows that $f \in R[a, c]$ and $f \in R[c, b]$ hence we'd simultaneously show that
$$\sup \left\{\ L(f, P_{[a, c]}) \ | \ P_{[a, c]} \text{ is a partition of } [a, c]\right\} + \sup\left\{\ L(f, P_{[c, b]}) \ | \ P_{[c, b]} \text{ is a partition of } [c, b]\right\} = \sup \left\{\ L(f, P_{[a, b]}) \ | \ P_{[a, b]} \text{ is a partition of } [a, b]\right\}$$ which would prove what we wanted to show.
I managed to show that $$\inf \left\{\ U(f, P_{[a, c]}) \ | \ P_{[a, c]} \text{ is a partition of } [a, c]\right\} + \inf \left\{\ U(f, P_{[c, b]}) \ | \ P_{[c, b]} \text{ is a partition of } [c, b]\right\} = \inf \left\{\ U(f, P') \ | P' = P_{[a, c]} \cup P_{[c, b]} \right\}$$
where $P_{[a, c]} \text{ is a partition of } [a, c]$ and $P_{[c, b]} \text{ is a partition of } [c, b]$. If I can show that $$\inf \left\{\ U(f, P') \ | P' = P_{[a, c]} \cup P_{[c, b]} \right\} = \inf \left\{\ U(f, P_{[a, b]}) \ | \ P_{[a, b]} \text{ is a partition of } [a, b]\right\}$$
But now I'm not sure how to proceed and show the above equality holds. How could I prove the above equality?
Also am I on the right track? Are there easier ways to prove this?
A different approach is shown here.
Following your approach, we first prove that if $f$ is integrable over $[a,b]$ then it is integrable over the subintervals.
By the Riemann condition, for any $\epsilon >0$ there exists a partition $P'_{[a,b]}$ such that
$$U(P'_{[a,b]},f) - L(P'_{[a,b]},f) < \epsilon$$
If the partition does not include the point $c$, then add it to form the partition $P_{[a,b]}$. This induces partitions of the subintervals such that $P_{[a,b]} = P_{[a,c]} \cup P_{[c,b]}.$ Furthermore, we now have
$$\tag{1}U(P_{[a,b]},f) = U(P_{[a,c]},f) + U(P_{[c,b]},f), \\ L(P_{[a,b]},f) = L(P_{[a,c]},f) + L(P_{[c,b]},f). $$
Since the new partition is a refinement we have
$$L(P'_{[a,b]},f) \leqslant L(P_{[a,b]},f) \leqslant U(P_{[a,b]},f) \leqslant U'(P_{[a,b]},f), $$
and it follows that
$$0 \leqslant U(P_{[a,c]},f)-L(P_{[a,c]},f)\leqslant U(P_{[a,b]},f)-L(P_{[a,b]},f) \leqslant U(P'_{[a,b]},f)-L(P'_{[a,b]},f) < \epsilon,$$
with a similar result for the Darboux sums over $[c,b]$. This proves integrability over the subintervals and gives us a starting point for equating the integrals.
The decomposition in (1) is true for any partitions where $P_{[a,b]} = P_{[a,c]} \cup P_{[c,b]}.$ If $P'_{[a,b]}$ is any partition with $P_{[a,b]}$ the refinement including the point $c$, we have (since the upper integral is the supremum of upper sums)
$$U(P'_{[a,b]},f) \leqslant U(P_{[a,b]},f) = U(P_{[a,c]},f) + U(P_{[c,b]},f) \leqslant \overline{\int_a^c }f + \overline{\int_c^b }f.$$
Taking the supremum of the LHS over partitions $P'_{[a,b]}$ we get
$$\tag{2}\overline{\int_a^b }f \leqslant \overline{\int_a^c }f + \overline{\int_c^b }f.$$
Taking arbitrary partitions $P'_{[a,c]}$ and $P''_{[c,b]}$ with $P'_{[a,c]} \cup P''_{[c,b]} = P_{[a,b]}$ we have
$$U(P'_{[a,c]},f) + U(P''_{[c,b]},f) = U(P_{[a,b]},f) \leqslant \overline{\int_a^b }f, $$
and taking suprema over partitions $P'_{[a,c]}$ and $P''_{[c,b]}$ it follows that
$$\tag{3} \overline{\int_a^c }f + \overline{\int_c^b }f \leqslant \overline{\int_a^b }f .$$
Together, (2) and (3) show that
$$\overline{\int_a^c }f + \overline{\int_c^b }f = \overline{\int_a^b }f,$$
and by a similar argument (or having already proved integrability over $[a,c]$ and $[c,b]$) we have
$$\underline{\int_a^c }f + \underline{\int_c^b }f = \underline{\int_a^b }f.$$
Therefore,
$$\int_a^c f + \int_c^b f = \int_a^b f.$$